Answer: Option B
$$\eqalign{
& {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} + {c^{\frac{1}{3}}} = 0 \cr
& \Rightarrow {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} = - {c^{\frac{1}{3}}} \cr
& {\text{Take cube on both sides}} \cr
& \Rightarrow {\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right)^3} = {\left( { - {c^{\frac{1}{3}}}} \right)^3} \cr
& \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right) = - c \cr
& \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }} - {c^{\frac{1}{3}}}} \right) = - c \cr
& \Rightarrow a + b + c = 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} \cr
& {\text{Again taking cube }} \cr
& \Rightarrow {\left( {a + b + c} \right)^3} = 27abc \cr} $$