Question

If $a+b+c=9,$ and $a^2+b^2+c^2=35,$ find thevalue of $a^3+b^3+c^3-3abc.$

Answer 1

Given,

$a+b+c=9,$ and $a^2+b^2+c^2=35,$

Using Identity,

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

$\therefore a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ac)$

$2ab+2bc+2ac=9^2-35$

$\therefore ab+bc+ac=23$

Again , Using Identity,

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

$=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$

$=(a+b+c)(a^2+b^2+c^2-(ab+bc+ac\left)\right)$

$a^3+b^3+c^3-3abc=\left(9\right)(35-23)$

$a^3+b^3+c^3-3abc=\left(9\right)\left(12\right)$

$\therefore a^3+b^3+c^3-3abc=108$