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 $x^{3}+y^{3}+z^{3}$ divisible by prime number is______

Answer 1

$( x + y + z )^{2}=144($ given $)$
$\Rightarrow \sum x ^{2}+2 \sum xy =144$
$\Rightarrow 96+2 \sum xy =144$
$ \Rightarrow \sum xy =24$
again $\frac{1}{ x }+\frac{1}{ y }+\frac{1}{ z }=36 $
$\Rightarrow xyz =\frac{24}{36}=\frac{2}{3}$
now $x ^{3}+ y ^{3}+ z ^{3}-3 xyz $
$=( x + y + z )\left(\sum x ^{2}-\sum xy \right)$
$\Rightarrow \sum x ^{3}-2=(12)(96-24)$
$=(12)(72)=864$
$\Rightarrow a x ^{3}=866$