Question

If $a+b+c=15$ and $a^2+b^2+c^2=83$ then find the value of $a^3+b^3+c^3-3abc$

Answer 1

Given: $a+b+c=15$ and $a^2+b^2+c^2=83$

Now, consider, $a+b+c=15$. On squaring both the sides, we get,

$\Rightarrow (a+b+c{)}^2={15}^2$

$\Rightarrow a^2+b^2+c^2+2(ab+bc+ac)=225$ [$\because (x+y+z{)}^2=x^2+y^2+z^2+2(xy+yz+zx)$]

$\Rightarrow 2(ab+bc+ca)=225-83$ [$\because a^2+b^2+c^2=83$ ]

$\Rightarrow ab+bc+ca=\frac{142}{2}=71$

Now, we know that $x^3+y^3+z^3-3xyz=(x+y+z)[x^2+y^2+z^2-(xy+yz+xz\left)\right]$

$\therefore a^3+b^3+c^3-3abc=(a+b+c)[a^2+b^2+c^2-(ab+bc+ca\left)\right]$

$\therefore a^3+b^3+c^3-3abc=\left(15\right)[83-(71\left)\right]$

$\Rightarrow a^3+b^3+c^3-3abc=15\times 12$

$\Rightarrow a^3+b^3+c^3-3abc=180$

Hence, $a^3+b^3+c^3-3abc=180$.