if a+b+c =5 and ab+bc+ca=10, then prove that a^{3}+b^{3}+c^{3}-3abc= -25.

a^{3}+b^{3}+c^{3}- 3abc = (a+b+c)*(a^{2}+b^{2}+c^{2}-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 a^{2}+b^{2}+c^{2}.

(a+b+c)*[a^{2}+b^{2}+c^{2}- ( ab+bc+ca) ] ( -ab-bc-ca can be written as - (ab+bc+ca, u must know this ofc )

Putting the values we get-

5* ( a^{2}+b^{2}+c^{2}- 10)

We need to find a^{2}+b^{2}+c^{2}, AND with values we have been given. If we think a little, we get this :-

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

5^{2} = a^{2}+b^{2}+c^{2}+ 2( ab+bc+ca) [ Taking 2 as common ]

25= a^{2}+b^{2}+c^{2} + 2(10)

25 = a^{2}+b^{2}+c^{2} + 20

Therefore, a^{2}+b^{2}+c^{2}= 5

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

5* ( 5-10)

= 5* (-5)

= -25.

Proved!

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a^{3}+b^{3}+c^{3}- 3abc = (a+b+c)*(a^{2}+b^{2}+c^{2}-ab-bc-ca)

a+b+c= 5 (given).

ab+ bc + ca = 10 (given)

5* ( a^{2}+b^{2}+c^{2}- 10)

We need to find a^{2}+b^{2}+c^{2}, AND with values we have been given. If we think a little, we get this :-

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

5^{2}= a^{2}+b^{2}+c^{2}+ 2( ab+bc+ca) [ Taking 2 as common ]

25= a^{2}+b^{2}+c^{2}+ 2(10)

25 = a^{2}+b^{2}+c^{2}+ 20

Therefore, a^{2}+b^{2}+c^{2}= 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!

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*The identity we need to prove is,*

a

a

^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)*We are provided only with (a+b+c) and ab+bc+ca, so we need to find a*

^{2}+b^{2}+c^{2}. In order to find a^{2}+b^{2}+c^{2}let us do (a+b+c)^{2}.*So,*

(a+b+c)

^{2}= a

^{2}+b

^{2}+c

^{2}+2ab+2bc+2ca

(a+b+c)

^{2}= a

^{2}+b

^{2}+c

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

*Now, let us substitute the values.*

(5)

^{2}= a

^{2}+b

^{2}+c

^{2}+2(10)

25

^{ }= a

^{2}+b

^{2}+c

^{2}+20

25-20 = a

^{2}+b

^{2}+c2

a

^{2}+b

^{2}+c

^{2}= 5

a

^{3}+b

^{3}+c

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

^{2}+b

^{2}+c

^{2}

^{}-ab-bc-ca)

-25 = 5(5-10)

= 5*-5

-25 = -25

Therefore, LHS=RHS

Hence Proved

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