Question

solve the factorisation given above

Answer 1

$<b>$Answer :

There are two ways to solve this question.

Let's try both, apply any of the one, that you feel easy to do!

$= > \frac{9 {x}^{3} {y}^{3} - 6 {x}^{3}y }{3xy} \\ \\taking \: a \: term\: common \\ \\ = > \frac{3 {x}^{3} y \: (3 {y}^{2} - 2) }{ 3xy} \\ \\ = > \frac{3xy \times {x}^{2} \: (3 {y}^{2} - 2)}{3xy} \\ \\ = > {x}^{2} \: (3 {y}^{2} - 2) \\ \\ = > 3 {x}^{2} {y}^{2} - 2 {x}^{2}$

Another way is :

$= > \frac{9 {x}^{3} {y}^{3} - 6 {x}^{3}y }{3xy} \\ \\ = > \frac{9 {x}^{3} {y}^{3} }{3xy} - \frac{6 {x}^{3} y}{3xy} \\ \\ = > 3 {x}^{2} {y}^{2} - 2 {x}^{2}$

$\color{red}\underline\textbf{Hence, the answer is same. }$

Tysm for the question!