Question

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

• A. $27abc$
• B. $3abc$
• C. $9abc$
• D. $abc$

Answer 1

The correct option is A $27abc$

Given, $\sqrt[3]{3a}+\sqrt[3]{3b}+\sqrt[3]{3c}=0$

i.e. $a^{1/3}+b^{1/3}+c^{1/3}=0$

$\Rightarrow a^{1/3}+b^{1/3}=-c^{1/3}$...............(1)

Cubing both sides, we get,

$\Rightarrow (a^{1/3}{)}^3+(b^{1/3}{)}^3+3a^{1/3}{b}^{1/3}(a^{1/3}+b^{1/3})=-c$ {Using $(a+b{)}^3=a^3+b^3+3ab(a+b)$}

$\Rightarrow a+b+3a^{1/3}{b}^{1/3}(-c^{1/3})=-c$ {Using (1)}

$\Rightarrow a+b+c=3a^{1/3}{b}^{1/3}{c}^{1/3}$

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

Therefore, option $A$ is correct.