Question:
Find the value
$8 x^{2} y^{3}-x^{5}$
Solution:
$=x^{2}\left((2 y)^{3}-x^{3}\right)$
$=x^{2}(2 y-x)\left((2 y)^{2}+2 y x x+x^{2}\right)$
$\left[\therefore x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)\right]$
$=x^{2}(2 y-x)\left(4 y^{2}+2 x y+x^{2}\right)$
$\therefore 8 x^{2} y^{3}-x^{5}=x^{2}(2 y-x)\left(4 y^{2}+2 x y+x^{2}\right)$