Question

If $a+b+c=8$ and $ab+bc+ca=20$, then the value of $a^3+b^3+c^3–3abc$ is ___

Answer 1

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

$a+b+c=8\text{and}ab+bc+ca=20$

${\left(8\right)}^2=a^2+b^2+c^2+2\times 20$

$64=a^2+b^2+c^2+40$

$a^2+b^2+c^2=24$

We now use the following identity:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca\left)\right)$

$a^3+b^3+c^3-3abc=8\times (24-20)=4\times 8=32$

Thus, $a^3+b^3+c^3-3abc=32$