Question

If $x+y+z=12$ and $x^2+y^2+z^2=96$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=36$ then the value of $x^3+y^3+z^3$ is

Answer 1

Given $x+y+z=12$ and $x^2+y^2+z^2=96$
Squaring it, we get
$(x+y+z{)}^2=12$
$x^2+y^2+z^2+2(xy+yz+zx)=144$
Therefore $xy+yz+zx=24$
[Since, given $x^2+y^2+z^2=96$]
Also $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=36$
Implies $\frac{(xy+yz+zx)}{xyz}=36$
$\Rightarrow xyz=\frac{(xy+yz+zx)}{36}$
$=\frac{24}{36}$
$\therefore xyz=\frac{2}{3}$

We know $(x^3+y^3+z^3)=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz$

$\Rightarrow x^3+y^3+z^3=\left(12\right)(96-24)+3\left(\frac{2}{3}\right)$

$=(12\times 72)+2$
$=864+2$
$=866$
Hence, the value of the required expression is $866$