(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3 (a-b)^3+(b-c)^3+(c-a)^3 =

MCQ

$\frac{(\text{a}^2-\text{b}^2)^3+(\text{b}^2-\text{c}^2)^3+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}=$

A
$3(a + b)( b+ c)(c + a)$
B
$3(a - b)(b - c)(c - a)$
C
$(a - b)(b - c)(c - a)$
✓
None of these.

✓

Answer

Correct option: D.

None of these.

If $a + b + c = 0$ then, $a^3 + b^3 + c^3 = 3abc$
Now, $(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2) = a^2 - b^2 + b^2 - c^2 + c^2 - a^2 = 0$
$\Rightarrow (a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3 = 3(a^2 - b^2)(b^2 - c^2)(c^2 - a^2)$
Again, $(a - b) + (b - c) + (c - a) = a - b + b - c + c - a = 0$
$\Rightarrow (a - b)^3 + (b - c)^3 + (c - a)^3 = 3(a - b)(b - c)(c - a)$
Thus, we have
$\frac{(\text{a}^2-\text{b}^2)^3+(\text{b}^2-\text{c}^2)^3+(\text{c}^2-\text{a}^2)^3}{(\text{a}-\text{b})^3+(\text{b}-\text{c})^3+(\text{c}-\text{a})^3}$
$=\frac{3(\text{a}^2-\text{b}^2)(\text{b}^2-\text{c}^2)(\text{c}^2-\text{a}^2)}{3(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$=\frac{(\text{a}-\text{b})(\text{a}+\text{b})(\text{b}-\text{c})(\text{b}+\text{c})(\text{c}-\text{a})(\text{c}+\text{a})}{(\text{a}-\text{b})(\text{b}-\text{c})(\text{c}-\text{a})}$
$=(\text{a}+\text{b})(\text{b}+\text{c})(\text{c}+\text{a})$
Hence, correct option is $(d).$