$(a - b)^{3} + (b - c)^{3} + (c - a)^{3}$ is equal to

3 a b c

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

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3 a^{3} b^{3} c^{3}

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[a - (b + c)]^{3}

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Solution

The correct option is B $3 (a - b) (b - c) (c - a)$
$(a - b^{3}) + (b - c^{3}) + {(c - a)}^{3} =$
$a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3} + b^{3} - 3 b^{2} c + 3 b c^{2} - c^{3} + c^{3} - 3 a c^{2} + 3 c a^{2} - a^{3}$
$= - 3 a^{2} b + 3 a b^{2} - 3 b^{2} c + 3 b c^{2} - 3 a c^{2} + 3 a^{2} c$
$= 3 a b (b - a) + 3 b c (c - b) + 3 a c (a - c)$
$= (3) (a - b) (b - c) (c - a)$

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Similar questions

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

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

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

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

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

{(a + b + c)}^{3} - a^{3} - b^{3} - c^{3} = [{(a + b + c)}^{3} - a^{3}] - [b^{3} + c^{3}]

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