Question

Factorise : 9 (x+y)²- 16 (x - 3y)²​

Answer 1

${\text{Let\:(x + y) be a \: \& \:(x - 3y)\:be\:b }}$

Now,

${9a}^{2} - {16b}^{2} \\ {(3a)}^{2} - {(4b)}^{2} \\ (3a - 4b)(3a + 4b)$

Putting the value of a & b in the term obtained above, we get,

${ \large{[3(x + y) - 4(x - 3y)][3(x + y) + 4(x - 3y)]}} \\ \\ { \large{[(3x + 3y - 4x + 12y)][(3x + 3y + 4x - 12y)]}} \\ \\ { \large{[ - x + 15y][7x - 9y]}} \\ \\ { \large{ { - 7x}^{2} + 9xy + 105xy - {135y}^{2} }} \\ \\ { \red{ \large{ \boxed{{ - 7x}^{2} + 114xy - {135y}^{2} }}}}$

Answer 2

Step-by-step explanation:

$9 {(x + y)}^{2} - 16 {(x - 3y)}^{2} \\ = {3}^{2} {(x + y)}^{2} - {4}^{2} {(x - 3y)}^{2} \\ = {(3x + 3y)}^{2} - {(4x - 12y)}^{2} \\ = (3x + 3y + 4x - 12y)(3x + 3y - 4x + 12y) \\ = (7x - 9y)(15y - x)$