Question
Factorize: $4(x - y)^2 - 12(x - y)(x + y) + 9(x + y)^2$
✓
Answer
$4(x - y)^2 - 12(x - y)(x + y) + 9(x + y)^{2$}$
$Let(x - y) = x,(x + y) = y = 4x^2 - 12xy + 9y^2$
Splitting the middle term $-12 = -6 - 6$
also $4 \times 9 = -6 \times -6 = 4x^2 - 6xy - 6xy + 9y^2$
$= 2x(2x - 3y) - 3y(2x - 3y)$
$= (2x - 3y)(2x - 3y) = (2x - 3y)^2$
Substituting $x = x - y y = x + y$
$= [2(x - y) - 3(x + y)]^2$
$= [2x - 2y - 3x - 3y]^2$
$= (2x - 3x - 2y - 3y)²$
$= [-x - 5y]^2 = [(-1)(x + 5y)]^2$
$= (x + 5y)^2 [(-1)^2 = 1]$
$\therefore$ $4(x - y)^2 - 12(x - y)(x + y) + 9(x + y)^2 = (x + 5y)^2$
$Let(x - y) = x,(x + y) = y = 4x^2 - 12xy + 9y^2$
Splitting the middle term $-12 = -6 - 6$
also $4 \times 9 = -6 \times -6 = 4x^2 - 6xy - 6xy + 9y^2$
$= 2x(2x - 3y) - 3y(2x - 3y)$
$= (2x - 3y)(2x - 3y) = (2x - 3y)^2$
Substituting $x = x - y y = x + y$
$= [2(x - y) - 3(x + y)]^2$
$= [2x - 2y - 3x - 3y]^2$
$= (2x - 3x - 2y - 3y)²$
$= [-x - 5y]^2 = [(-1)(x + 5y)]^2$
$= (x + 5y)^2 [(-1)^2 = 1]$
$\therefore$ $4(x - y)^2 - 12(x - y)(x + y) + 9(x + y)^2 = (x + 5y)^2$
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