MCQ
$\frac{(a-b)^3-(a+b)^3}{2}+a\left(a^2+3 b^2\right)=$
- A$a^3-b^3$
- B$(a+b)^3$
- C$a^3+b^3$
- ✓$(a-b)^3$
✓
Answer
Correct option: D.
$(a-b)^3$
(d)
$\frac{(a-b)^3-(a+b)^3}{2}+a\left(a^2+3 b^2\right)$
$=\frac{\left(a^3-b^3-3 a^2 b+3 a b^2\right)-\left(a^3+b^3+3 a^2 b+3 a b^2\right)}{2}+a^3+3 a b^2$
$=-b^3-3 a^2 b+a^3+3 a b^2=(a-b)^3$
$\frac{(a-b)^3-(a+b)^3}{2}+a\left(a^2+3 b^2\right)$
$=\frac{\left(a^3-b^3-3 a^2 b+3 a b^2\right)-\left(a^3+b^3+3 a^2 b+3 a b^2\right)}{2}+a^3+3 a b^2$
$=-b^3-3 a^2 b+a^3+3 a b^2=(a-b)^3$
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