Find the value

Solution:

Let $(x-y)=x,(x+y)=y$

$=4 x^{2}-12 x y+9 y^{2}$

Splitting the middle term - 12 = – 6 - 6  also 4 × 9 = −6 × − 6

$=4 x^{2}-6 x y-6 x y+9 y^{2}$

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

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

 

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

Substituting x = x - y & y = x + y

$=[2(x-y)-3(x+y)]^{2}=[2 x-2 y-3 x-3 y]^{2}$

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

$=[-x-5 y]^{2}$

$=[(-1)(x+5 y)]^{2}$

$=(x+5 y)^{2} \quad\left[?(-1)^{2}=1\right]$

 

$\therefore 4(x-y)^{2}-12(x-y)(x+y)+9(x+y)^{2}=(x+5 y)^{2}$