The expression (a-b)^3+(b-c)^3+(c-a)^3 can be factorized as:

MCQ

The expression $(a-b)^3+(b-c)^3+(c-a)^3$ can be factorized as:

A
$(a-b)(b-c)(c-a)$
✓
$3(a-b)(b-c)(c-a)$
C
$-3(a-b)(b-c)(c-a)$
D
$(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$

✓

Answer

Correct option: B.

$3(a-b)(b-c)(c-a)$

By we know that $a^3+b^3+c^3-3 a b c$
$=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
If $a+b+c=0$, then
$a^3+b^3+c^3=3 a b c$
In given expression,
Let $a - b = A , b - c = B , c - a = C$
Now, $a-b+b-c+c-a=0$
i.e. $A+B+C=0$
$\Rightarrow A^3+B^3+C^3=3 A B C$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3$
$=3(a-b)(b-c)(c-a)$
Hence, correct option is $(b).$

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