MCQ
$(x+y)^3-(x-y)^3$ can be factorized as:
- ✓$2 y\left(3 x^2+y^2\right)$
- B$2 x\left(3 x^2+y^2\right)$
- C$2 y\left(3 y^2+x^2\right)$
- D$2 x\left(x^2+3 y^2\right)$
✓
Answer
Correct option: A.
$2 y\left(3 x^2+y^2\right)$
We know the identity
$a^3-b^3=(a-b)\left(a^2+b^2+a b\right)$
Let $x+y=a$ and $x-y=b$
Then,
$a^3-b^3$
$=(x+y)^3-(x-y)^3$
$=[(x+y)-(x-y)]\left[(x+y)^2+(x-y)^2+(x+y)(x-y)\right]$
$=2 y\left[x^2+y^2+2 x y+x^2+y^2-2 x y+x^2-y^2\right]$
$=2 y\left(3 x^2+y^2\right)$
Hence, correct option is $(a).$
$a^3-b^3=(a-b)\left(a^2+b^2+a b\right)$
Let $x+y=a$ and $x-y=b$
Then,
$a^3-b^3$
$=(x+y)^3-(x-y)^3$
$=[(x+y)-(x-y)]\left[(x+y)^2+(x-y)^2+(x+y)(x-y)\right]$
$=2 y\left[x^2+y^2+2 x y+x^2+y^2-2 x y+x^2-y^2\right]$
$=2 y\left(3 x^2+y^2\right)$
Hence, correct option is $(a).$
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