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Find the value

Question:

Find the value

$8 x^{2} y^{3}-x^{5}$

Solution:

$=x^{2}\left((2 y)^{3}-x^{3}\right)$

$=x^{2}(2 y-x)\left((2 y)^{2}+2 y x x+x^{2}\right)$

$\left[\therefore x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)\right]$

$=x^{2}(2 y-x)\left(4 y^{2}+2 x y+x^{2}\right)$

$\therefore 8 x^{2} y^{3}-x^{5}=x^{2}(2 y-x)\left(4 y^{2}+2 x y+x^{2}\right)$

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