Question:
Find the value
$4(x-y)^{2}-12(x-y)(x+y)+9(x+y)^{2}$
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
$=[2(x-y)-3(x+y)]^{2}=[2 x-2 y-3 x-3 y]^{2}$
$=(2 x-3 x-2 y-3 y)^{2}$
$=[-x-5 y]^{2}$
$=[(-1)(x+5 y)]^{2}$
$=(x+5 y)^{2} \quad\left[?(-1)^{2}=1\right]$
$\therefore 4(x-y)^{2}-12(x-y)(x+y)+9(x+y)^{2}=(x+5 y)^{2}$
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