Question

If $a+b+c=8$ and $ab+bc+ca=20$, find the value of $a^3+b^3+c^3-3abc$

Answer 1

Since $(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$
$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$
$a+b+c=8$ and $ab+bc+ca=20$,
$(8{)}^2=a^2+b^2+c^2+2\times (20)$
$\Rightarrow 64=a^2+b^2+c^2+40$
$\therefore a^2+b^2+c^2=64-40=24$
We know that
$a^3+b^3+c^3-3abc$
$=(a+b+c)\{a^2+b^2+c^2-(ab+bc+ca\left)\right\}$
$\therefore a^3+b^3+c^3-3abc$
$=8\times (24-20)=4\times 8=32$
$[\because a+b+c=8,ab+bc+ca=20$ and $a^2+b^2+c^2=24]$
Thus, $a^3+b^3+c^3-3abc=32$