(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, $a^3+b^3+c^3=3 a b c$
$\text { Now, }\left(a^2-b^2\right)+\left(b^2-c^2\right)+\left(c^2-a^2\right)=a^2-b^2+b^2-c^2+c^2-a^2=0$
$\Rightarrow\left(a^2-b^2\right)^3+\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3=3\left(a^2-b^2\right)\left(b^2-c^2\right)\left(c^2-a^2\right)$
$\text { Again, }(a-b)+(b-c)+(c-a)=a-b+b-c+c-a=0$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)$
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 $(d).$

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