Rd Sharma 2019 Solutions for Class 9 Math Chapter 4 Algebraic Identities are provided here with simple step-by-step explanations. These solutions for Algebraic Identities are extremely popular among Class 9 students for Math Algebraic Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 Book of Class 9 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2019 Solutions. All Rd Sharma 2019 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Page No 4.11:

#### Question 1:

Write the following in the expanded form:

(i) (*a +*2*b + c*)^{2}

(ii) (2*a* − 3*b* − *c*)^{2}

(iii) (−3*x + y + z*)^{2}

(iv) (*m* + 2*n* − 5*p*)^{2}

(v) (2* + x* − 2*y*)^{2}

(vi) (*a*^{2} + *b*^{2} + *c*^{2})^{2}

(vii) (*ab + bc + ca*)^{2}

(viii) ${\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)}^{2}$

(ix) ${\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)}^{2}$

(x) $(x+2y+4z{)}^{2}$

(xi) (2*x − y + z*)^{2}

(xii) (−2*x* + 3*y* + 2*z*)^{2}

#### Answer:

In the given problem, we have to find expended form

(i) Given

We shall use the identity

Here

By applying in identity we get

Hence the expended form of is

(ii) Given

We shall use the identity

Here

By applying in identity we get

Hence the expended form of is

(iii) Given

We shall use the identity ${\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca$

Here

By applying in identity we get

${\left(-3x+y+z\right)}^{2}={\left(-3x\right)}^{2}+{y}^{2}+{z}^{2}-2\times 3x\times y+2yz-2\times \left(-3x\right)\times z$

Hence the expended form of is.

(iv) Given

We shall use the identity

Here

By applying in identity we get

Hence the expended form of is

(v) Given

We shall use the identity ${\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca$

Here

By applying in identity we get

Hence the expended form of is.

(vi) Given

We shall use the identity

Here

By applying in identity we get

Hence the expended form of is.

(vii) Given

We shall use the identity

Here

By applying in identity we get

Hence the expended form ofis .

(viii) Given

We shall use the identity

Here

By applying in identity we get

Hence the expended form of is

(ix) Given

We shall use the identity

Here

By applying in identity we get

Hence the expended form of is

(x) Given

We shall use the identity ${\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca$

Here

By applying in identity we get

Hence the expended form of is

(xi) Given

We shall use the identity ${\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca$

Here

By applying in identity we get

Hence the expended form of is

(xii) Given

We shall use the identity ${\left(a+b+c\right)}^{2}={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca$

Here

By applying in identity we get

Hence the expended form of is.

#### Page No 4.12:

#### Question 2:

If *a + b + c* = 0 and *a*^{2} + *b*^{2} + *c*^{2} = 16, find the value of *ab + bc + ca.*

#### Answer:

In the given problem, we have to find value of

Given and

Squaring the equation, we get

Now putting the value of in above equation we get,

Taking 2 as common factor we get

Hence the value of is .

#### Page No 4.12:

#### Question 3:

If *a*^{2} + *b*^{2} + *c*^{2} = 16 and *ab* + *bc* + *ca* = 10, find the value of *a + b + c*.

#### Answer:

In the given problem, we have to find value of

Given

Multiply equation with 2 on both sides we get,

Now adding both equation and we get

We shall use the identity

Hence the value of is .

#### Page No 4.12:

#### Question 4:

If *a + b + c* = 9 and *ab + bc + ca* = 23, find the value of *a*^{2} + *b*^{2}^{ }+ *c*^{2}.

#### Answer:

In the given problem, we have to find value of

Given

Squaring both sides of we get,

Substituting in above equation we get,

Hence the value of is.

#### Page No 4.12:

#### Question 5:

Find the value of 4x^{2} + y^{2} + 25x^{2 }+ 4xy − 10yz − 20zx when x= 4, y = 3 and z = 2.

#### Answer:

In the given problem, we have to find value of

Given

We have

This equation can also be written as

Using the identity

${\left(x+y-z\right)}^{2}={x}^{2}+{y}^{2}+{z}^{2}+2xy-2yz-2xz$

Hence the value of is .

#### Page No 4.12:

#### Question 6:

Simplify:

(i) (*a +b + c*)^{2} + (*a − b + c*)^{2}

(ii) (*a +b + c*)^{2} − (*a − b + c*)2

(iii) (*a +b + c*)^{2 }+ (*a − b + c*)^{2} + (*a +b − c*)^{2}

(iv) (2*x + p − c*)^{2 }− (2*x − p + c*)^{2}

(v) (*x*^{2} + *y*^{2 }− *z*^{2}) − (*x*^{2} − *y*^{2} + *z*^{2})^{2}

#### Answer:

In the given problem, we have to simplify the expressions

(i) Given

By using identity

Hence the equation becomes

Taking 2 as common factor we get

Hence the simplified value of is

(ii) Given

By using identity

Hence the equation becomes

Taking 4 as common factor we get

Hence the simplified value of is.

(iii) Given

By using identity , we have

Taking 3 as a common factor we get

Hence the value ofis

.

(iv) Given

By using identity , we get

By cancelling the opposite terms, we get

Taking as common a factor we get,

Hence the value of is

(v) We have (*x*^{2} + *y*^{2 }− *z*^{2}) − (*x*^{2} − *y*^{2} + *z*^{2})^{2}

Using formula, we get

(*x*^{2} + *y*^{2 }− *z*^{2}) − (*x*^{2} − *y*^{2} + *z*^{2})^{2
}

By canceling the opposite terms, we get

Taking as common factor we get

Hence the value of is.

#### Page No 4.12:

#### Question 7:

Simplify each of the following expressions:

(i) $(x+y+z{)}^{2}+{\left(x+\frac{y}{2}+\frac{z}{3}\right)}^{2}-{\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}\right)}^{2}$

(ii) (x + y− 2z)^{2} − x^{2} − y^{2} − 3z^{2} + 4xy

(iii) (x^{2}− x + 1)^{2} − (x^{2} + x + 1)^{2}

#### Answer:

In the given problem, we have to simplify the value of each expression

(i) Given

We shall use the identity for each bracket

$-\left\{{\left(\frac{x}{2}\right)}^{2}+{\left(\frac{y}{3}\right)}^{2}+{\left(\frac{z}{4}\right)}^{2}+2\left(\frac{x}{2}\right)\left(\frac{y}{3}\right)+2\left(\frac{y}{3}\right)\left(\frac{z}{4}\right)+2\left(\frac{x}{2}\right)\left(\frac{z}{4}\right)\right\}$

By arranging the like terms we get

Now adding or subtracting like terms,

Hence the value of is

(ii) Given

We shall use the identity ${\left(x+y-z\right)}^{2}={x}^{2}+{y}^{2}+{z}^{2}+2xy-2yz-2zx$ for expanding the brackets

Now arranging liked terms we get,

Hence the value of is

(iii) Given

We shall use the identity for each brackets

${\left({x}^{2}-x+1\right)}^{2}-{\left({x}^{2}+x+1\right)}^{2}=\left[{\left({x}^{2}\right)}^{2}+{\left(-x\right)}^{2}+{1}^{2}-2{x}^{3}-2x+2{x}^{2}\right]\phantom{\rule{0ex}{0ex}}-\left[{\left({x}^{2}\right)}^{2}+{\left(x\right)}^{2}+{1}^{2}+2{x}^{3}+2x+2{x}^{2}\right]$

Canceling the opposite term and simplifies

Hence the value of is .

#### Page No 4.19:

#### Question 1:

Find the cube of each of the following binomials expressions:

(i) $\frac{1}{x}+\frac{y}{3}$

(ii) $\frac{3}{x}-\frac{2}{{x}^{2}}$

(iii) $2x+\frac{3}{x}$

(iv) $4-\frac{1}{3x}$

#### Answer:

In the given problem, we have to find cube of the binomial expressions

(i) Given

We shall use the identity

Here

By applying the identity we get

Hence cube of the binomial expression is

(ii) Given

We shall use the identity

Here

By applying the identity we get

Hence cube of the binomial expression of is

(iii) Given

We shall use the identity .

Here,

By applying identity we get

Hence cube of the binomial expression of is

(iv) Given

We shall use the identity

Here

By applying in identity we get

Hence cube of the binomial expression of is .

#### Page No 4.19:

#### Question 2:

If a + b = 10 and *ab* = 21, find the value of a^{3} + b^{3}

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Hence the value of is .

#### Page No 4.19:

#### Question 3:

If a − b = 4 and *ab* = 21, find the value of a^{3} −b^{3}

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Hence the value of is .

#### Page No 4.20:

#### Question 4:

If $x+\frac{1}{x}=5$, find the value of ${x}^{3}+\frac{1}{{x}^{3}}$

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Hence the value of is

#### Page No 4.20:

#### Question 5:

If $x-\frac{1}{x}=7$, find the value of ${x}^{3}-\frac{1}{{x}^{3}}$

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Hence the value of is

#### Page No 4.20:

#### Question 6:

If $x-\frac{1}{x}=5$, find the value of ${x}^{3}-\frac{1}{{x}^{3}}$

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Hence the value of is .

#### Page No 4.20:

#### Question 7:

If ${x}^{2}+\frac{1}{{x}^{2}}$ = 51, find the value of ${x}^{3}-\frac{1}{{x}^{3}}$

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

In order to find we are using identity

Here and

Hence the value of is .

#### Page No 4.20:

#### Question 8:

If ${x}^{2}+\frac{1}{{x}^{2}}=98$, find the value of ${x}^{3}+\frac{1}{{x}^{3}}$

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

In order to find we are using identity

Here and

Hence the value of is .

#### Page No 4.20:

#### Question 9:

If 2x+3y = 13 and xy = 6, find the value of 8x^{3} + 27y^{3}

#### Answer:

In the given problem, we have to find the value of

Given,

In order to find we are using identity

Here putting,

Hence the value of is .

#### Page No 4.20:

#### Question 10:

If 3x − 2y = 11 and xy = 12, find the value of 27x^{3} − 8y^{3}

#### Answer:

In the given problem, we have to find the value of

Given,

In order to find we are using identity

Here putting,,

Hence the value of is.

#### Page No 4.20:

#### Question 11:

Evaluate each of the following:

(i) (103)^{3}

(ii) (98)^{3}

(iii) (9.9)^{3}

(iv) (10.4)3

(v) (598)^{3}

(vi) (99)^{3}

#### Answer:

In the given problem, we have to find the value of numbers

(i) Given

In order to find we are using identity

We can write as

Hence where

The value of is

(ii) Given

In order to find we are using identity

We can write as

Hence where

The value of is

(iii) Given

In order to find we are using identity

We can write as

Hence where

The value of is

(iv) Given

In order to find we are using identity

We can write as

Hence where

The value of is

(v) Given

In order to find we are using identity

We can write as

Hence where

The value of is

(vi) Given

In order to find we are using identity

We can write as

Hence where

The value of is .

#### Page No 4.20:

#### Question 12:

Evaluate each of the following:

(i) 111^{3} − 89^{3}

(ii) 46^{3}+34^{3}

(iii) 104^{3} + 96^{3}

(iv) 93^{3} − 107^{3}

#### Answer:

In the given problem, we have to find the value of numbers

(i) Given

We can write as

We shall use the identity

Here

${111}^{3}-{89}^{3}={\left(100+11\right)}^{3}-{\left(100-11\right)}^{3}$

Hence the value of is

(ii) Given

We can write as

We shall use the identity

Here

${46}^{3}+{34}^{3}={\left(40+6\right)}^{3}+{\left(40-6\right)}^{3}$

Hence the value of is

(iii) Given

We can write as

We shall use the identity

Here

${104}^{3}+{96}^{3}={\left(100+4\right)}^{3}+{\left(100-4\right)}^{3}$

Hence the value of is

(iv) Given

We can write as

We shall use the identity

Here

Hence the value of is .

#### Page No 4.20:

#### Question 13:

If $x+\frac{1}{x}=3$, calculate ${x}^{2}+\frac{1}{{x}^{2}},{x}^{3}+\frac{1}{{x}^{3}}$ and ${x}^{4}+\frac{1}{{x}^{4}}$

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Again squaring on both sides we get,

We shall use the identity

Again cubing on both sides we get,

We shall use identity

Hence the value of is respectively.

#### Page No 4.20:

#### Question 14:

Find the value of 27x^{3} + 8y^{3}, if

(i) 3x + 2y = 14 and xy = 8

(ii) 3x + 2y = 20 and xy = $\frac{14}{9}$

#### Answer:

In the given problem, we have to find the value of

(i) Given

On cubing both sides we get,

We shall use identity

Hence the value of is

(ii) Given

On cubing both sides we get,

We shall use identity

Hence the value of is .

#### Page No 4.20:

#### Question 15:

Find the value of 64x^{3} − 125z^{3}, if 4x − 5z = 16 and xz = 12.

#### Answer:

From given problem we have to find the value of

Given

On cubing both sides of we get

We shall use identity

Hence the value of is .

#### Page No 4.20:

#### Question 16:

If $x-\frac{1}{x}=3+2\sqrt{2}$, find the value of ${x}^{3}-\frac{1}{{x}^{3}}$

#### Answer:

In the given problem, we have to find the value of

Given

Cubing on both sides of we get

${\left(x-\frac{1}{x}\right)}^{3}={\left(3+2\sqrt{2}\right)}^{3}$

We shall use identity

$27+16\sqrt{2}+18\sqrt{2}\times 3+18\sqrt{2}\times 2\sqrt{2}={x}^{3}-\frac{1}{{x}^{3}}-9-6\sqrt{2}\phantom{\rule{0ex}{0ex}}27+16\sqrt{2}+54\sqrt{2}+72={x}^{3}-\frac{1}{{x}^{3}}-9-6\sqrt{2}$

Hence the value of is .

#### Page No 4.20:

#### Question 17:

Simplify each of the following:

(i) (x+3)^{3} + (x−3)^{3}

(ii) ${\left(\frac{x}{2}+\frac{y}{3}\right)}^{3}-{\left(\frac{x}{2}-\frac{y}{3}\right)}^{3}$

(iii) ${\left(x+\frac{2}{x}\right)}^{3}+{\left(x-\frac{2}{x}\right)}^{3}$

(iv) (2x − 5y)^{3} − (2x + 5y)^{3}

#### Answer:

In the given problem, we have to simplify equation

(i) Given

We shall use the identity

Here

By applying identity we get

Hence simplified form of expression is .

(ii) Given

We shall use the identity

Here

By applying identity we get

By rearranging the variable we get

Hence the simplified value of is

(iii) Given

We shall use the identity

Here

By applying identity we get

By rearranging the variable we get,

Hence the simplified value of is

(iv) Given

We shall use the identity

Here

By applying the identity we get

By rearranging the variable we get,

Hence the simplified value of is .

#### Page No 4.20:

#### Question 18:

If ${x}^{4}+\frac{1}{{x}^{4}}=194,$find ${x}^{3}+\frac{1}{{x}^{3}},{x}^{2}+\frac{1}{{x}^{2}}$and $x+\frac{1}{x}$

#### Answer:

In the given problem, we have to find the value of

Given

By adding and subtracting in left hand side of we get,

Again by adding and subtracting in left hand side of we get,

Now cubing on both sides of we get

we shall use identity

Hence the value of is respectively.

#### Page No 4.20:

#### Question 19:

If ${x}^{4}+\frac{1}{{x}^{4}}=119$, find the value of ${x}^{3}-\frac{1}{{x}^{3}}$

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

In order to find we are using identity

${\left(x-\frac{1}{x}\right)}^{2}={x}^{2}+\frac{1}{{x}^{2}}-2\times x\times \frac{1}{x}\phantom{\rule{0ex}{0ex}}$

In order to find we are using identity

Here and

Hence the value of is .

#### Page No 4.24:

#### Question 1:

Find the following products:

(i) (3*x* + 2*y*) (9x^{2} − 6xy + 4y^{2})

(ii) (4*x* − 5*y*) (16x^{2} + 20xy + 25y^{2})

(iii) (7p^{4} + q) (49p^{8} − 7p^{4}q + q^{2})

(iv) $\left(\frac{x}{2}+2y\right)\left(\frac{{x}^{2}}{4}-xy+4{y}^{2}\right)\phantom{\rule{0ex}{0ex}}$

(v) $\left(\frac{3}{x}-\frac{5}{y}\right)\left(\frac{9}{{x}^{2}}+\frac{25}{{y}^{2}}+\frac{15}{xy}\right)$

(vi) $\left(3+\frac{5}{x}\right)\left(9-\frac{15}{x}+\frac{25}{{x}^{2}}\right)$

(vii) $\left(\frac{2}{x}+3x\right)\left(\frac{4}{{x}^{2}}+9{x}^{2}-6\right)$

(viii) $\left(\frac{3}{x}-2{x}^{2}\right)\left(\frac{9}{{x}^{2}}+4{x}^{4}-6x\right)$

(ix) (1 − *x*) (1+ *x* *+ x*^{2})

(x) (1 + *x*) (1 − *x + x*^{2})

(xi) (*x*^{2} − 1) (*x*^{4} + *x*^{2} + 1)

(xii) (*x*^{3} + 1) (*x*^{6} − *x*^{3 + }1)

#### Answer:

(i) In the given problem, we have to find the value of

Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(ii) Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(iii) Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(iv) Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(v) Given

We shall use the identity

We can rearrange the as

$=\left(\frac{3}{x}\right)\times \left(\frac{3}{x}\right)\times \left(\frac{3}{x}\right)-\left(\frac{5}{y}\right)\times \left(\frac{5}{y}\right)\times \left(\frac{5}{y}\right)\phantom{\rule{0ex}{0ex}}=\frac{27}{{x}^{3}}-\frac{125}{{y}^{3}}$

Hence the Product value of is

(vi) Given

We shall use the identity ,

we can rearrange the as

Hence the Product value of is

(vii) Given

We shall use the identity,

We can rearrange the as

Hence the Product value of is

(viii) Given

We shall use the identity

We can rearrange the as

$\left(\frac{3}{x}-2{x}^{2}\right)\left({\left(\frac{3}{x}\right)}^{2}+{\left(2{x}^{2}\right)}^{2}-\left(\frac{3}{x}\right)\left(2{x}^{2}\right)\right)\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{x}\right)}^{3}-{\left(2{x}^{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}=\left(\frac{3}{x}\right)\left(\frac{3}{x}\right)\left(\frac{3}{x}\right)-\left(2{x}^{2}\right)\left(2{x}^{2}\right)\left(2{x}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{27}{{x}^{3}}-8{x}^{6}$

Hence the Product value of is

(ix) Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(x) Given

We shall use the identity

We can rearrange the as

Hence the Product value of is

(xi) Given

We shall use the identity

We can rearrange the as

$\left({x}^{2}-1\right)\left[{\left({x}^{2}\right)}^{2}+\left({x}^{2}\right)\left(1\right)+{\left(1\right)}^{2}\right]$

Hence the Product value of is

(xii) Given

We shall use the identity,

We can rearrange the as

Hence the Product value of is .

#### Page No 4.24:

#### Question 2:

If *x* = 3 and *y* = − 1, find the values of each of the following using in identify:

(i) (9*y*^{2 }− 4*x*^{2}) (81*y*^{4} +36*x*^{2}*y*^{2} + 16*x*^{4})

(ii) $\left(\frac{3}{x}-\frac{x}{3}\right)\left(\frac{{x}^{2}}{9}+\frac{9}{{x}^{2}}+1\right)$

(iii) $\left(\frac{x}{7}+\frac{y}{3}\right)\left(\frac{{x}^{2}}{49}+\frac{{y}^{2}}{9}-\frac{xy}{21}\right)$

(iv) $\left(\frac{x}{y}-\frac{y}{3}\right)\frac{{x}^{2}}{16}+\frac{xy}{12}+\frac{{y}^{2}}{9}$

(v) $\left(\frac{5}{x}+5x\right)$ $\left(\frac{25}{{x}^{2}}-25+25{x}^{2}\right)$

#### Answer:

In the given problem, we have to find the value of equation using identity

(i) Given

We shall use the identity

We can rearrange the as

Now substituting the value in we get,

Hence the Product value of is

(ii) Given

We shall use the identity

We can rearrange the as

$=\left(\frac{3}{x}\right)\times \left(\frac{3}{x}\right)\times \left(\frac{3}{x}\right)-\left(\frac{x}{3}\right)\times \left(\frac{x}{3}\right)\times \left(\frac{x}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{27}{{x}^{3}}-\frac{{x}^{3}}{27}$

Now substituting the value in we get,

Hence the Product value of is

(iii) Given

We shall use the identity,

We can rearrange the as

Now substituting the value in

Taking Least common multiple, we get

Hence the Product value of is

(iv) Given

We shall use the identity

We can rearrange the as

$=\left(\frac{x}{4}\right)\times \left(\frac{x}{4}\right)\times \left(\frac{x}{4}\right)-\left(\frac{y}{3}\right)\times \left(\frac{y}{3}\right)\times \left(\frac{y}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{{x}^{3}}{64}-\frac{{y}^{3}}{27}$

Now substituting the value in we get,

Taking Least common multiple, we get

Hence the Product value of is

(v) Given

We shall use the identity,

We can rearrange the as

Now substituting the value in

Taking Least common multiple, we get

Hence the Product value of is .

#### Page No 4.25:

#### Question 3:

If *a + b* = 10 and *ab* = 16, find the value of *a*^{2} − *ab + b*^{2} and *a*^{2} + *ab* + *b*^{2}

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity ${\left(a+b\right)}^{3}={a}^{3}+{b}^{3}+3ab(a+b)$

We can rearrange the identity as

Now substituting values in as,

We can write as

Now rearrange as

Thus

Now substituting values

Hence the value of is respectively.

#### Page No 4.25:

#### Question 4:

If *a + b* = 8 and *ab* = 6, find the value of *a*^{3} + *b*^{3}

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Hence the value of is .

#### Page No 4.25:

#### Question 5:

If *a + b* = 6 and *ab* = 20, find the value of *a*^{3} − *b*^{3}

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Hence the value of is .

#### Page No 4.25:

#### Question 6:

If *x* = −2 and *y* = 1, by using an identity find the value of the following

(i) 4*y*^{2 − }9*x*^{2} (16*y*^{4} + 36*x*^{2}*y*^{2}+81*x*^{4})

(ii) $\left(\frac{2}{x}-\frac{x}{2}\right)\left(\frac{4}{{x}^{2}}+\frac{{x}^{2}}{4}+1\right)$

(iii) $\left(5y+\frac{15}{y}\right)\left(25{y}^{2}-75+\frac{225}{{y}^{2}}\right)$

#### Answer:

(i) In the given problem, we have to find the value of using identity

Given

We shall use the identity

We can rearrange the as

$=\left(4{y}^{2}\right)\times \left(4{y}^{2}\right)\times \left(4{y}^{2}\right)-\left(9{x}^{2}\right)\times \left(9{x}^{2}\right)\times \left(9{x}^{2}\right)\phantom{\rule{0ex}{0ex}}=64{y}^{6}-729{x}^{6}$

Now substituting the value in we get,

Taking 64 as common factor in above equation we get,

Hence the Product value of is

(ii) In the given problem, we have to find the value of using identity

Given

We shall use the identity

We can rearrange the as

$=\left(\frac{2}{x}\right)\times \left(\frac{2}{x}\right)\times \left(\frac{2}{x}\right)-\left(\frac{x}{2}\right)\times \left(\frac{x}{2}\right)\times \left(\frac{x}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{8}{{x}^{3}}-\frac{{x}^{3}}{8}$

Now substituting the value in we get,

Hence the Product value of is = 0.

(iii) Given

We shall use the identity,

We can rearrange the as

Now substituting the value in

Hence the Product value of is

#### Page No 4.28:

#### Question 1:

Find the following products:

(i) (3x + 2y + 2z) (9x^{2} + 4y^{2} + 4z^{2} − 6xy − 4yz − 6zx)

(ii) (4x − 3y + 2z) (16x^{2} + 9y^{2} + 4z^{2} + 12xy + 6yz − 8zx)

(iii) (2ab − 3b − 2c) (4a^{2} + 9b^{2} +4c^{2} + 6 ab − 6 bc + 4ca)

(iv) (3x − 4y + 5z) (9x^{2} +16y^{2}^{ }+ 25z^{2} + 12xy −15zx + 20yz)

#### Answer:

In the given problem, we have to find Product of equations

(i)Given

We shall use the identity

Hence the product of is

(ii) Given

We shall use the identity

Hence the product of is

(iii) Given

We shall use the identity

Hence the product of is

(iv) Given

We shall use the identity

Hence the product of is

#### Page No 4.29:

#### Question 2:

Evaluate:

(i) 25^{3} − 75^{3} + 50^{3}

(ii) 48^{3} − 30^{3} − 18^{3}

(iii) ${\left(\frac{1}{2}\right)}^{3}+{\left(\frac{1}{3}\right)}^{3}-{\left(\frac{5}{6}\right)}^{3}$

(iv) (0.2)^{3} − (0.3)^{3} + (0.1)^{3}

#### Answer:

In the given problem we have to evaluate the following

(i) Given

We shall use the identity

Let Take

Hence the value of is

(ii) Given

We shall use the identity

Let Take

Hence the value of is

(iii) Given

We shall use the identity

Let Take

Applying least common multiple we get,

Hence the value of is

(iv) Given

We shall use the identity

Let Take

${a}^{3}+{b}^{3}+{c}^{3}=\left(0.2-0.3+0.1\right)\left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca\right)+3abc$

${a}^{3}+{b}^{3}+{c}^{3}=0\times \left({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca\right)+3abc$

Hence the value of is

#### Page No 4.29:

#### Question 3:

If x + y + z = 8 and xy +yz +zx = 20, find the value of x^{3} + y^{3}^{ }+ z^{3} −3xyz

#### Answer:

In the given problem, we have to find value of

Given

We shall use the identity

We know that

Here substituting we get

Hence the value of is .

#### Page No 4.29:

#### Question 4:

If *a* + *b* + *c* = 9 and *ab* +*bc* + *ca* = 26, find the value of a^{3} + b^{3}+ c^{3} − 3*abc*

#### Answer:

In the given problem, we have to find value of

Given

We shall use the identity

We know that

Here substituting we get,

Hence the value of is .

#### Page No 4.29:

#### Question 5:

If *a + b + c* = 9 and *a*^{2}+ *b*^{2} + *c*^{2} =35, find the value of *a*^{3} + *b*^{3 }+ c^{3} −3*abc*

#### Answer:

In the given problem, we have to find value of

Given

We shall use the identity

We know that

Here substituting we get

Hence the value of is .

#### Page No 4.29:

#### Question 1:

If x + $\frac{1}{x}$= 3, then find the value of ${x}^{2}+\frac{1}{{x}^{2}}$.

#### Answer:

We have to find the value of

Given

Using identity

Here

By substituting the value of we get,

By transposing + 2 to left hand side, we get

Hence the value of is .

#### Page No 4.29:

#### Question 2:

If $x+\frac{1}{x}=3$, then find the value of ${x}^{6}+\frac{1}{{x}^{6}}$.

#### Answer:

We have to find the value of

Given

Using identity

Here

By substituting the value of We get,

By transposing + 2 to left hand side, we get

Cubing on both sides we get,

Using identity ${\left(a+b\right)}^{3}={a}^{3}+{b}^{3}+3ab\left(a+b\right)$

Here

Put we get

By transposing 21 to left hand side we get ,

Hence the value of is .

#### Page No 4.29:

#### Question 3:

If a + b = 7 and ab = 12, find the value of a^{2} + b^{2}

#### Answer:

We have to find the value of

Given

Using identity

By substituting the value of we get

By transposing +24 to left hand side we get ,

Hence the value of is .

#### Page No 4.29:

#### Question 4:

If *a* −* b* = 5 and *ab* = 12, find the value of a^{2} + b^{2}.

#### Answer:

We have to find the value

Given

Using identity

By substituting the value of we get ,

By transposing – 24 to left hand side we get

Hence the value of is .

#### Page No 4.29:

#### Question 5:

If $x-\frac{1}{x}=\frac{1}{2}$, then write the value of $4{x}^{2}+\frac{4}{{x}^{2}}$

#### Answer:

We have to find the value of

Given

Using identity

Here

By substituting the value of we get

By transposing – 2 to left hand side we get

By taking least common multiply we get

By multiplying 4 on both sides we get

Hence the value of is

#### Page No 4.29:

#### Question 6:

If ${a}^{2}+\frac{1}{{a}^{2}}=102$, find the value of $a-\frac{1}{a}$.

#### Answer:

We have to find the value of

Given

Using identity

Here

By substituting we get

Hence the value of is

#### Page No 4.29:

#### Question 7:

If a + b + c = 0, then write the value of $\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}$

#### Answer:

We have to find the value of

Given

Using identity

Put

Hence the value of is

#### Page No 4.30:

#### Question 1:

Mark the correct alternative in each of the following:

(1) If $x+\frac{1}{x}=5$, then ${x}^{2}+\frac{1}{{x}^{2}}=\phantom{\rule{0ex}{0ex}}$

(a) 25

(b) 10

(c) 23

(d) 27

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here put,

Hence the value of is

Hence the correct choice is (c).

#### Page No 4.30:

#### Question 2:

If $x+\frac{1}{x}=2$, then ${x}^{3}+\frac{1}{{x}^{3}}=$

(a) 64

(b) 14

(c) 8

(d) 2

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Hence the value of is

Hence the correct choice is (d).

#### Page No 4.30:

#### Question 3:

If $x+\frac{1}{x}$ = 4, then ${x}^{4}+\frac{1}{{x}^{4}}=$

(a) 196

(b) 194

(c) 192

(d) 190

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here put,

Squaring on both sides we get,

Hence the value of is

Hence the correct choice is (b).

#### Page No 4.30:

#### Question 4:

If $x+\frac{1}{x}=3$, then ${x}^{6}+\frac{1}{{x}^{6}}$ =

(a) 927

(b) 414

(c) 364

(d) 322

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identityand

Here put,

Take Cube on both sides we get,

Hence the value of is

Hence the correct choice is (d).

#### Page No 4.30:

#### Question 5:

If ${x}^{2}+\frac{1}{{x}^{2}}=102$, then $x-\frac{1}{x}$ =

(a) 8

(b) 10

(c) 12

(d) 13

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Hence the value of is

Hence the correct choice is (b).

#### Page No 4.30:

#### Question 6:

If ${x}^{3}+\frac{1}{{x}^{3}}=110$, then $x+\frac{1}{x}=$

(a) 5

(b) 10

(c) 15

(d) none of these

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Put we get,

Substitute y = 5 in the above equation we get

The Equation satisfy the condition that

Hence the value of is 5

The correct choice is (a).

#### Page No 4.30:

#### Question 7:

If ${x}^{3}-\frac{1}{{x}^{3}}=14$, then $x-\frac{1}{x}=$

(a) 5

(b) 4

(c) 3

(d) 2

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Put we get,

Substitute *y* = 2 in above equation we get,

The Equation satisfy the condition that

Hence the value of is 2

Hence the correct choice is (d).

#### Page No 4.30:

#### Question 8:

If *a* + *b* + *c* = 9 and *ab* + *bc* + *ca* = 23, then *a ^{2}* +

*b*+

^{2}*c*

^{2}=

(a) 35

(b) 58

(c) 127

(d) none of these

#### Answer:

We have to find

Given

Using identity we get,

By transposing +46 to left hand side we get,

Hence the value of is

The correct choice is (a).

#### Page No 4.30:

#### Question 9:

(a − b)^{3} + (b − c)^{3} + (c − a)^{3} =

(a) (a + b + c) (a^{2} + b^{2}^{ }+ c^{2} − ab − bc − ca)

(b) (a − b) (b − c) (c − a)

(c) 3(a − b) ( b− c) (c − a)

(d) none of these

#### Answer:

Given

Using identity

Here

Hence the Value of is

The correct choice is .

#### Page No 4.30:

#### Question 10:

If $\frac{a}{b}+\frac{b}{a}=-1$, then a^{3} − b^{3} =

(a) 1

(b) −1

(c) $\frac{1}{2}$

(d) 0

#### Answer:

Given

Taking Least common multiple in we get,

Using identity

Hence the value of is

The correct choice is (d).

#### Page No 4.30:

#### Question 11:

If a − b = −8 and ab = −12, then a^{3} − b^{3} =

(a) −244

(b) −240

(c) −224

(d) −260

#### Answer:

To find the value of a^{3} − b^{3}

Given

Using identity

Here we get

Transposing -288 to left hand side we get

Hence the value of is -224

The correct choice is .

#### Page No 4.31:

#### Question 12:

If the volume of a cuboid is 3x^{2} − 27, then its possible dimensions are

(a) 3, x^{2}, − 27x

(b) 3, x − 3, x + 3

(c) 3, x^{2}, 27x

(d) 3, 3, 3

#### Answer:

We have to find the possible dimension of cuboid

Given: volume of cuboid

Take 3 as common factor

Using identity

We get,

Here the dimension of cuboid is 3,

The correct alternate is .

#### Page No 4.31:

#### Question 13:

75 × 75 + 2 × 75 × 25 + 25 × 25 is equal to

(a) 10000

(b) 6250

(c) 7500

(d) 3750

#### Answer:

We have to find the product of

Using identity

Here

Hence the product of is 10,000

The correct choice is .

#### Page No 4.31:

#### Question 14:

(x − y) (x + y) (x^{2} + y^{2}) (x^{4}^{ }+ y^{4}) is equal to

(a) x^{16} − y^{16}

(b) x^{8} − y^{8}

(c) x^{8} + y8

(d) x^{16} + y^{16}

#### Answer:

Given

Using the identity

Hence is equal to

The correct choice is .

#### Page No 4.31:

#### Question 15:

If ${x}^{4}+\frac{1}{{x}^{4}}=623$, then $x+\frac{1}{x}=$

(a) 27

(b) 25

(c) $3\sqrt{3}$

(d) $-3\sqrt{3}$

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here put,

We shall use the identitywe get,

Taking square root on both sides we get,

Hence the value of is

Hence the correct choice is (c).

#### Page No 4.31:

#### Question 16:

If ${x}^{4}+\frac{1}{{x}^{4}}=194,$ then ${x}^{3}+\frac{1}{{x}^{3}}=$

(a) 76

(b) 52

(c) 64

(d) none of these

#### Answer:

Given

Using identity

Here,

Again using identity

Here

Substituting

Using identity

Here

Hence the value of is

The correct choice is (b).

#### Page No 4.31:

#### Question 17:

If $x-\frac{1}{x}=\frac{15}{4}$, then $x+\frac{1}{x}$=

(a) 4

(b) $\frac{17}{4}$

(c) $\frac{13}{4}$

(d) $\frac{1}{4}$

#### Answer:

In the given problem, we have to find the value of

Given

We shall use the identity

Here putting,

Substitute in we get,

Hence the value of is

Hence the correct choice is (b).

#### Page No 4.31:

#### Question 18:

If $3x+\frac{2}{x}=7$, then $\left(9{x}^{2}-\frac{4}{{x}^{2}}\right)=$

(a) 25

(b) 35

(c) 49

(d) 30

#### Answer:

We have to find the value of

Given

Using identity we get,

Here

Substituting we get,

By transposing left hand side we get,

Again using identity we get,

Substituting we get

Using identity we get

Here

Substituting we get,

The value of is

The correct choice is (b)

#### Page No 4.31:

#### Question 19:

If *a*^{2} + *b*^{2} + *c*^{2} − *ab* − *bc* − *ca* =0, then

(a) *a* + *b* + *c*

(b) *b* + *c* = a

(c) c + a = b

(d) a = b = c

#### Answer:

Given

Multiplying both sides by 2 we get,

Therefore the sum of positive quantities is zero if and only if each quantity is zero.

If, then

The correct choice is (d).

#### Page No 4.31:

#### Question 20:

If *a* + *b* + *c* = 0, then $\frac{{a}^{2}}{bc}+\frac{{b}^{2}}{ca}+\frac{{c}^{2}}{ab}=$

(a) 0

(b) 1

(c) −1

(d) 3

#### Answer:

We have to find

Given

Using identity

Hence the value of

The correct choice is (d).

#### Page No 4.31:

#### Question 21:

If *a*^{1}^{/3} + *b*^{1}^{/3} + *c*^{1}^{/3} = 0, then

(a) a + *b* + *c* = 0

(b) (*a* + *b* + *c*)^{3} =27*abc*

(c) *a* + *b* + *c* = 3*abc*

(d) *a*^{3} + *b*^{3} + *c*^{3} = 0

#### Answer:

Given

Using identity we get

Here

Taking Cube on both sides we get,

Hence the value of is

The correct choice is .

#### Page No 4.31:

#### Question 22:

If *a* + *b* + *c* = 9 and *ab* + *bc* + *ca* =23, then *a*^{3}^{ }+ *b*^{3}^{ }+ *c*^{3} − 3*abc* =

(a) 108

(b) 207

(c) 669

(d) 729

#### Answer:

We have to find the value of

Given

Using identity we get,

By transposing +46 to left hand side we get,

Using identity

The value of is

Hence the correct choice is .

#### Page No 4.31:

#### Question 23:

$\frac{({a}^{2}-{b}^{2}{)}^{3}+({b}^{2}-{c}^{2}{)}^{3}+({c}^{2}-{a}^{2}{)}^{3}}{(a-b{)}^{3}+(b-c{)}^{3}+(c-a{)}^{3}}=$

(a) 3(a + b) ( b+ c) (c + a)

(b) 3(a − b) (b − c) (c − a)

(c) (a − b) (b − c) (c − a)

(d) none of these

#### Answer:

We have to find the value of

Using Identity we get,

Hence the value of is

The correct choice is .

#### Page No 4.32:

#### Question 24:

The product (a + b) (a − b) (a^{2} − ab + b^{2}) (a^{2} + ab + b^{2}) is equal to

(a) a^{6} + b^{6}

(b) a^{6} − b^{6}

(c) a^{3} − b^{3}

(d) a^{3}^{ }+ b^{3}

#### Answer:

We have to find the product of

Using identity

We can rearrange as

Again using the identity

Here

Hence the product of is

The correct choice is ** **.

#### Page No 4.32:

#### Question 25:

The product (x^{2}−1) (x^{4} + x^{2} + 1) is equal to

(a) x^{8} − 1

(b) x^{8} + 1

(c) x^{6} − 1

(d) x^{6} + 1

#### Answer:

We have to find the product of

Using identity

Here

Hence the product value of is

The correct alternate is .

#### Page No 4.32:

#### Question 26:

If $\frac{a}{b}+\frac{b}{a}=1$, then a^{3} + b^{3} =

(a) 1

(b) −1

(c) $\frac{1}{2}$

(d) 0

#### Answer:

Given

Using identity we get,

Hence the value of is .

The correct choice is (d).

#### Page No 4.32:

#### Question 27:

If 49a^{2} − b = $\left(7a+\frac{1}{2}\right)\left(7a-\frac{1}{2}\right)$, then the value of *b* is

(a) 0

(b) $\frac{1}{4}\phantom{\rule{0ex}{0ex}}$

(c) $\frac{1}{\sqrt{2}}$

(d) $\frac{1}{2}$

#### Answer:

We have to find the value of *b*

Given

Using identity

We get

Equating ‘*b*’ on both sides we get

Hence the value of *b* is

The correct choice is .

#### Page No 4.6:

#### Question 1:

Evaluate each of the following using identities:

(i) ${\left(2x-\frac{1}{x}\right)}^{2}$

(ii) (2*x + y*) (2*x − y*)

(iii) (*a*^{2}*b *- *b*^{2}*a*)^{2}

(iv) (*a - *0.1) (*a *+ 0.1)

(v) (1.5*x*^{2 }− 0.3*y*^{2}) (1.5*x*^{2 }+ 0.3*y*^{2})

#### Answer:

In the given problem, we have to evaluate expressions by using identities.

(i) Given

We shall use the identity

By applying identity we get

Hence the value of is

(ii) We have been given

We shall use the identity

Here,,

By applying identity we get

Hence the value ofis

(iii) The given expression is

We shall use the identity

Here

By applying identity we get

Hence the value of is

(iv) The given expression is $\left(a+0.1\right)\left(a-0.1\right)$

We shall use the identity

Here

By applying identity we get

$\left(a+0.1\right)\left(a-0.1\right)={\left(a\right)}^{2}-{\left(0.1\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(a\times a\right)-\left(0.1\times 0.1\right)\phantom{\rule{0ex}{0ex}}={a}^{2}-0.01$

Hence the value of$\left(a+0.1\right)\left(a-0.1\right)$is

(v) The given expression is

We shall use the identity

Here

By applying identity we get

Hence the value ofis

#### Page No 4.7:

#### Question 2:

Evaluate each of the following using identities:

(i) (399)^{2}

(ii) (0.98)^{2}

(iii) 991 ☓ 1009

(iv) 117 ☓ 83

#### Answer:

In the given problem, we have to evaluate expressions by using identities.

(i) Given

We can write as

We shall use the Identity

Where ,

By applying in identity we get

${\left(400-1\right)}^{2}={\left(400\right)}^{2}-2\times 400\times 1+{\left(1\right)}^{2}\phantom{\rule{0ex}{0ex}}=400\times 400-800+1\phantom{\rule{0ex}{0ex}}=160000-800+1\phantom{\rule{0ex}{0ex}}=159201$

Hence the value of is

(ii) We have been given

We can write as

We shall use the identity

Where,

By applying in identity we get

Hence the value of is

(iii) The given expression is

We have

So we can express and in the terms of as

We shall use the identity $\left(x-y\right)\left(x+y\right)={x}^{2}-{y}^{2}$

Here

By applying in identity we get

Hence the value of is

(iv) The given expression is

We have

So we can express and in the terms of 100 as

We shall use the identity $\left(x-y\right)\left(x+y\right)={x}^{2}-{y}^{2}$

Here

By applying in identity we get

Hence the value of is

#### Page No 4.7:

#### Question 3:

Simplify each of the following:

(i) 175 × 175 + 2 × 175 × 25 + 25 × 25

(ii) 322 × 322 − 2 × 322 × 22 + 22 × 22

(iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × × 0.24

(iv) $\frac{7.83\times 7.83-1.17\times 1.17}{6.66}$

#### Answer:

In the given problem, we have to simplify expressions

(i) Given

Put

Hence the equation becomes,

That is

Hence the value of is

(ii) We have been given

Put

Hence the equation becomes

That is

Hence the value of is

(iii) Given

Put

Hence the equation becomes

That is

Hence the value of is

(iv) We have been given

Put

Hence the equation becomes

Hence the value of is

#### Page No 4.7:

#### Question 4:

If $x+\frac{1}{x}=11,$ find the value of ${x}^{2}+\frac{1}{{x}^{2}}$.

#### Answer:

In the given problem, we have to find

Given

On squaring both sides we get,

Hence the value of is .

#### Page No 4.7:

#### Question 5:

If $x-\frac{1}{x}=-1$, find the value of ${x}^{2}+\frac{1}{{x}^{2}}$

#### Answer:

In the given problem, we have to find

Given

On squaring both sides we get,

We shall use the identity

Hence the value of is.

#### Page No 4.7:

#### Question 6:

If $x+\frac{1}{x}=\sqrt{5}$, find the value of ${x}^{2}+\frac{1}{{x}^{2}}$ and ${x}^{4}+\frac{1}{{x}^{4}}$.

#### Answer:

In the given problem, we have to find and

We have

On squaring both sides we get,

We shall use the identity

Again squaring on both sides we get,

We shall use the identity

Hence the value ofis and is .

#### Page No 4.7:

#### Question 7:

If 9*x*^{2 }+ 25*y*^{2} = 181 and *xy* = −6, find the value of 3*x* + 5*y.*

#### Answer:

In the given problem, we have to find

We have been given and

Let us take

We shall use the identity

By substituting and we get,

Hence the value of is.

#### Page No 4.7:

#### Question 8:

If 2*x* + 3*y* = 8 and *xy* = 2, find the value of 4*x*^{2} + 9*y*^{2}

#### Answer:

In the given problem, we have to find

We have been given and

Let us take

On squaring both sides we get,

We shall use the identity

By substituting we get,

Hence the value of is

#### Page No 4.7:

#### Question 9:

If 3*x* − 7*y* = 10 and *xy = *-1, find the value of 9*x*^{2} + 49*y*^{2}

#### Answer:

In the given problem, we have to find

We have been given and

Let us take

On squaring both sides we get,

We shall use the identity

By substituting we get,

Hence the value of is.

#### Page No 4.7:

#### Question 10:

Simplify each of the following products:

(i) $\left(\frac{1}{2}a-3b\right)\left(3b+\frac{1}{2}a\right)\left(\frac{1}{4}{a}^{2}+9{b}^{2}\right)$

(ii) ${\left(m+\frac{n}{7}\right)}^{3}\left(m-\frac{n}{7}\right)$

#### Answer:

(i) In the given problem, we have to find product of

We have been given

On rearranging we get,

We shall use the identity

By substituting,we get,

We shall use the identity

Hence the value of is

(ii) In the given problem, we have to find product of

We have been given

On rearranging we get

We shall use the identity

By substituting,, we get ,

Hence the value of is .

#### Page No 4.7:

#### Question 11:

If ${x}^{2}+\frac{1}{{x}^{2}}=66$, find the value of $x-\frac{1}{x}$

#### Answer:

In the given problem, we have to find

Given

Adding and subtracting 2 on left hand side

Hence the value of is

#### Page No 4.7:

#### Question 12:

If ${x}^{2}+\frac{1}{{x}^{2}}=79$, find the value of $x+\frac{1}{x}$

#### Answer:

In the given problem, we have to find

Given

Adding and subtracting 2 on left hand side,

Hence the value of is

#### Page No 4.7:

#### Question 13:

Simplify each of the following products:

(i) $\left(\frac{x}{2}-\frac{2}{5}\right)\left(\frac{2}{5}-\frac{x}{2}\right)-{x}^{2}+2x$

(ii) $\left({x}^{2}+x-2\right)\left({x}^{2}-x+2\right)$

(iii) $\left({x}^{3}-3{x}^{2}-x\right)\left({x}^{2}-3x+1\right)$

(iv) $\left(2{x}^{4}-4{x}^{2}+1\right)(2{x}^{4}-4{x}^{2}-1)$

#### Answer:

(i) In the given problem, we have to find product of

On rearranging we get

We shall use the identity

By substituting ,

Hence the value of is

(ii) In the given problem, we have to find product of

On rearranging we get

We shall use the identity

Hence the value of is

(iii) In the given problem, we have to find product of

Taking as common factor

We shall use the identity

Hence the value of is

(iv) In the given problem, we have to find product of

On rearranging we get

We shall use the identity

Hence the value of is.

#### Page No 4.7:

#### Question 14:

Prove that *a*^{2} + *b*^{2} + *c*^{2} −*ab−bc−ca *is always non-negative for all values of *a, b*, and *c*.

#### Answer:

In the given problem, we have to prove is always non negative for all that is we have to prove that

Consider,

Hence is always non negative for all

Note: Square of all negative numbers is always positive or non negative.

View NCERT Solutions for all chapters of Class 9