# if a+b+c =5 and ab+bc+ca=10, then prove that a3+b3+c3-3abc= -25.

regards

• 4

a3+b3+c3- 3abc = (a+b+c)*(a2+b2+c2-ab-bc-ca) [It is an identity]

a+b+c= 5 (given).

ab+ bc + ca = 10 (given)

Now concentrate a bit, if you look carefully at the identity, we have got 2 values, we need to find a2+b2+c2.

(a+b+c)*[a2+b2+c2- ( ab+bc+ca) ] ( -ab-bc-ca can be written as - (ab+bc+ca, u must know this ofc )

Putting the values we get-

5* ( a2+b2+c2- 10)

We need to find a2+b2+c2, AND with values we have been given. If we think a little, we get this :-

(a+b+c)2 = a2+b2=c2 + 2ab+ 2bc+ 2ca

52 = a2+b2+c2+ 2( ab+bc+ca) [ Taking 2 as common ]

25= a2+b2+c2 + 2(10)

25 = a2+b2+c2 + 20

Therefore, a2+b2+c2= 5

We got all values! letz put them in the equation we got :-

5* ( 5-10)

= 5* (-5)

= -25.

Proved!

• 76

a3+b3+c3- 3abc = (a+b+c)*(a2+b2+c2-ab-bc-ca)

a+b+c= 5 (given).

ab+ bc + ca = 10 (given)

5* ( a2+b2+c2- 10)

We need to find a2+b2+c2, AND with values we have been given. If we think a little, we get this :-

(a+b+c)2= a2+b2=c2+ 2ab+ 2bc+ 2ca

52= a2+b2+c2+ 2( ab+bc+ca) [ Taking 2 as common ]

25= a2+b2+c2+ 2(10)

25 = a2+b2+c2+ 20

Therefore, a2+b2+c2= 5

We got all values! put them in the equation we got :-

5* ( 5-10)

= 5* (-5)

= -25.

Hence The Proof!

We got all values! letz put them in the equation we got :-

= -25.

Proved!

• 8
The identity we need to prove is,
​a3+b3+c3-3abc = (a+b+c)(a2+b2+c2-ab-bc-ca)

We are provided only with (a+b+c) and ab+bc+ca, so we need to find a2+b2+c2. In order to find a2+b2+c2 let us do (a+b+c)2.
So,
(a+b+c)2 = a2+b2+c2+2ab+2bc+2ca
​(a+b+c)2 = a2+b2+c2+2(ab+bc+ca)
Now, let us substitute the values.
(5)2 = a2+b2+c2+2(10)
25 = a2+b2+c2+20
25-20 = a2+b2+c2
a2+b2+c2 = 5

a3+b3+c​3-3abc = (a+b+c)(a2+b2+c2-ab-bc-ca)
-25 = 5(5-10)
= 5*-5
-25 = -25
Therefore, LHS=RHS
Hence Proved

• 13
omg im only in 7th y is all dis coming wen i searched class 7
• -6
letz nhi lets
• -5