Question

If $a+b+c=5$ and $ab+bc+ca=10$, then prove that $a^3+b^3+c^3-3abc=-25$

Answer 1

We have,

$a+b+c=5$

$ab+bc+ca=10$

Since,

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

$a^2+b^2+c^2+2(ab+bc+ca)=25$

$a^2+b^2+c^2+2\times 10=25$

$a^2+b^2+c^2=5$

We know that,

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

$a^3+b^3+c^3-3abc=\left(5\right)\times (5-10)$

$a^3+b^3+c^3-3abc=5\times -5$

$a^3+b^3+c^3-3abc=-25$

Hence, proved.