Question

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

Answer 1

To prove, $a^3+b^3+c^3-3abc=-25$,
Given,
$a+b+c=5$, $ab+bc+ca=10$
$\because (a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$
$\therefore (5{)}^2=a^2+b^2+c^2+2(10)$
$\Rightarrow 25=a^2+b^2+c^2+20$
$\Rightarrow a^2+b^2+c^2=25–20$
$\Rightarrow a^2+b^2+c^2=5$
$LHS=a^3+b^3+c^3-3abc$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$=\left(5\right)[5-(ab+bc+ca\left)\right]$
$=5(5–10)=5(-5)=-25=RHS$
Hence proved.