The factors of $(a - b)^{3} + (b - c)^{3} + (c - a)^{3}$ are

(a + b + c) (a^{2} + b^{2} + c^{2} + a b + b c + a c)

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(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)

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

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(a - b) (b - c) (c - a)

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Solution

The correct option is D $3 (a - b) (b - c) (c - a)$
Recall that if $x + y + z = 0$ then $x^{3} + y^{3} + z^{3} = 3 x y z$
Here, $x = a - b, y = b - c, z = c - a$ and $x + y + z$ is obviously $= 0$
Thus,
$(a - b)^{3} + (b - c)^{3} + (c - a)^{3} = x^{3} + y^{3} + z^{3} = 3 x y z = 3 (a - b) (b - c) (c - a)$

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Similar questions

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

∣ ∣ ∣ ∣ \begin{matrix} a & b - c & c + b a + c & b & c - a a - b & b + a & c \end{matrix} ∣ ∣ ∣ ∣ = (a + b + c) (a^{2} + b^{2} + c^{2})

∣ ∣ ∣ ∣ \begin{matrix} a & b - c & c + b a + c & b & c - a a - b & b + a & c \end{matrix} ∣ ∣ ∣ ∣ = (a + b + c) (a^{2} + b^{2} + c^{2}) .

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