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

Answer: 2

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