If a + b + c = 0, then ( $a^{3}$ + $b^{3}$ + $c^{3}$ ) is equal to-

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abc

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3abc

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(ab + bc + ca)

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Solution

The correct option is C
3abc

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

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

a + b + c = 0 ⇒ $a^{3}$ + $b^{3}$ + $c^{3}$ - 3abc = 0

$a^{3}$ + $b^{3}$ + $c^{3}$ = 3abc

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Similar questions

a + b + c = 0

(a^{3} + b^{3} + c^{3}) \div a b c

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

> 0

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

a + b + c = 0

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

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