Question

If $a+b+c=9$, and $ab+bc+ac=23$, then $a^3+b^3+c^3-3abc=$

Answer 1

Given: $a+b+c=9$ and $ab+bc+ac=23$
As, $a+b+c=9$
Squaring on both sides, we get
$(a+b+c{)}^2=9^2$
$a^2+b^2+c^2+2ab+2bc+2ac=81$
$a^2+b^2+c^2+2(ab+bc+ac)=81$
$a^2+b^2+c^2+2\left(23\right)=81$
$a^2+b^2+c^2+46=81$
$a^2+b^2+c^2=81-46$
$a^2+b^2+c^2=35\dots \left(1\right)$
Using identity ,
$a^3+b^3+c^3-3abc$
$=(a+b+c)[a^2+b^2+c^2-(ab+bc+ac\left)\right]$
$\therefore a^3+b^3+c^3-3abc=9(35-23)$
$=108$

Hence, $\left(a\right)$ is the correct option.