Question

If a + b + c = 15 and a² + b + c = 83, find the value of a + b3 + c3-3abc.​

Answer 1

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Answer:

180

Step-by-step explanation:

Given,

(a+b+c)=15

(a^2+b^2+c^2)=83

We now that,

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

15^2 = 83 + 2(ab+bc+ca)

225 - 83 = 2(ab+bc+ca)

142=2(ab+bc+ca)

\implies ab+bc+ca = 71

Also, we know that,

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

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

=(15)(83 - 71)

=15\times 12

=180

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