Factorize:$ x^{3}-2 x^{2} y+3 x y^{2}-6 y^{3} $

Given :

\( x^{3}-2 x^{2} y+3 x y^{2}-6 y^{3} \)

To do :

We have to factorize the given expression.

Solution :

$x^3 - 2x^2y + 3xy^2 - 6y^3 = x^2(x - 2y) + 3y^2(x - 2y)$

$= (x - 2y) (x^2 + 3y^2)$

Hence, $x^3 - 2x^2y + 3xy^2 - 6y^3 =(x - 2y) (x^2 + 3y^2)$.