If a3+b3+c3=0, then (a+b+c)3=

a+b+c=0⇒a3+b3+c3=3abc∴a3+b3+c3=0⇒(a3)3+(b3)3+(c3)3=3.a3.b3.c3⇒a+b+x=3(abc3)⇒(a+b+c)3=27abc

If a3+b3+c3=0, then (a+b+c)3=

a+b+c=0⇒a3+b3+c3=3abc∴a3+b3+c3=0⇒(a3)3+(b3)3+(c3)3=3.a3.b3.c3⇒a+b+x=3(abc3)⇒(a+b+c)3=27abc

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If $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$ , then ${(a + b + c)}^{3} =$

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$9 a b c$

$a b c$

$3 a b c$

$27 a b c$

answer is D.

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Detailed Solution

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

$∴ \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0 \Rightarrow {(\sqrt[3]{a})}^{3} + {(\sqrt[3]{b})}^{3} + {(\sqrt[3]{c})}^{3} = 3. \sqrt[3]{a} . \sqrt[3]{b} . \sqrt[3]{c}$

$\Rightarrow a + b + x = 3 (\sqrt[3]{a b c})$

$\Rightarrow {(a + b + c)}^{3} = 27 a b c$

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If $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$ , then ${(a + b + c)}^{3} =$