Question : If $(x+y)^3+8(x-y)^3=(3 x+A y)\left(3 x^2+B x y+C y^2\right)$, then the value of (A + B + C) is:
Option 1: 0
Option 2: 4
Option 3: 2
Option 4: 3
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Correct Answer: 0
Solution :
$(x+y)^3+8(x-y)^3=(3 x+A y)\left(3 x^2+B x y+C y^2\right)$
Taking L.H.S.
$(x+y)^3+8(x-y)^3$
$=(x+y)^3+[2(x-y)]^3$
Using identity, $a^3+b^3=(a+b)(a^2-ab+b^2)$
$=[x+y+2(x-y)][(x+y)^2-2(x+y)(x-y)+[2(x-y)]^2]$
$=(3x-y)[(x^2+y^2+2xy) -2(x^2-y^2)+[4(x^2+y^2-2xy]]$
$=(3x-y)[x^2+y^2+2xy -2x^2+2y^2+4x^2+4y^2-8xy]$
$=(3x-y)(3x^2 -6xy+7y^2)$
On comparing the coefficient with R.H.S., we get,
⇒ A = –1
⇒ B = –6
⇒ C = 7
$\therefore$ (A + B + C) = – 1 – 6 + 7 = 0
Hence, the correct answer is 0.
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