MCQ
If $(x+y)^3-(x-y)^3-6 y\left(x^2-y^2\right)=k y^2$, then $k=$
- A$1$
- B$2$
- C$4$
- ✓$8$
✓
Answer
Correct option: D.
$8$
Let $x+y=A$ and $x-y=B$
Now, $(A-B)^3=A^3-B^3-3 A B(A-B)$
$\Rightarrow[(x+y)-(x-y)]^3=(x+y)^3-(x-y)^3-3(x+y)(x-y)[(x+y)-(x-y)]$
$=(x+y)^3-(x-y)^3-3\left(x^2-y^2\right)(2 y)$
$=(x+y)^3-(x-y)^3-6 y\left(x^2-y^2\right)$
But, $(x+y)^3-(x-y)^3-6 y\left(x^2-y^2\right)=k y^3$
$\Rightarrow[(x+y)-(x-y)]^3=(2 y)^3=k 8 y^3$
$\Rightarrow(2 y)^3=k y^3$
$\Rightarrow 8 y^3=k y^3$
$\Rightarrow k=8$
Hence, correct option is $(d).$
Now, $(A-B)^3=A^3-B^3-3 A B(A-B)$
$\Rightarrow[(x+y)-(x-y)]^3=(x+y)^3-(x-y)^3-3(x+y)(x-y)[(x+y)-(x-y)]$
$=(x+y)^3-(x-y)^3-3\left(x^2-y^2\right)(2 y)$
$=(x+y)^3-(x-y)^3-6 y\left(x^2-y^2\right)$
But, $(x+y)^3-(x-y)^3-6 y\left(x^2-y^2\right)=k y^3$
$\Rightarrow[(x+y)-(x-y)]^3=(2 y)^3=k 8 y^3$
$\Rightarrow(2 y)^3=k y^3$
$\Rightarrow 8 y^3=k y^3$
$\Rightarrow k=8$
Hence, correct option is $(d).$
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