Question

If a + b + c = 15 and a2 + b2 + c2 = 83, find the value of a3 + b3 + c3 – 3abc.

Answer 1

I hope this answer help you

Answer 2

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,

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

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

$=(15)(83 - 71)$

$=15\times 12$

$=180$