If $a + b + c = 0$ , then what is the value of ${(a + b + c)}^{3} + {(c + a - b)}^{3} + {(b + c - a)}^{3}$

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

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

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24 a b c

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- 24 a b c

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Solution

The correct option is D $- 24 a b c$
$a + b + c = 0$ $\Rightarrow$ $a + b = - c, c + a = - b$
$b + c = a$
$∴ {(a + b - c)}^{3} + {(c + a - b)}^{3} + {(b + c - a)}^{3}$
$= {(- c - c)}^{3} + {(- b - b)}^{3} + {(- a - a)}^{3}$
$= {(- 2 c)}^{3} + {(- 2 b)}^{3} + {(- 2 a)}^{3}$
$= - 8 c^{3} - 8 b^{3} - 8 a^{3} = - 8 (a^{3} + b^{3} + c^{3})$
$= - 8 \times 3 a b c = - 24 a b c$

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Similar questions

a + b + c = 0

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

If $a + b + c = 0$ , then the value of $(a + b - c)^{3} + (b + c - a)^{3} + (c + a - b)^{3}$ is:

> 0

\frac{a^{3}}{b^{3}} + \frac{b^{3}}{c^{3}} + \frac{c^{3}}{a^{3}} \geq 3

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

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

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