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

0

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a b c

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3 a b c

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2 a b c

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Solution

The correct option is B $3 a b c$
Given , $a + b + c = 0$
Multiply both sides by $a^{2} + b^{2} + c^{2} - a b - b c - a c$
$(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c) = 0$
We know that,
$(a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c) =$ $a^{3} + b^{3} + c^{3} - 3 a b c$

$\Rightarrow$ $a^{3} + b^{3} + c^{3} - 3 a b c = 0$
$\Rightarrow$ $a^{3} + b^{3} + c^{3} = 3 a b c$

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Similar questions

If a+b+c=0 then $(a^{3} + b^{3} + c^{3})$ is
(a) 0 (b) abc (c) 2abc (d) 3abc

a + b + c = 0

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

a + b + c = 0

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

a + b + c = 0

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

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