Answer: Option D
$$\eqalign{
& \root 3 \of a + \root 3 \of b + \root 3 \of c \cr
& {\text{Take cube both sides}} \cr
& {\left( {\root 3 \of a + \root 3 \of b } \right)^3} = {\left( {\root 3 \of c } \right)^3} \cr
& \Rightarrow a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {\root 3 \of a + \root 3 \of b } \right) = c \cr
& \Rightarrow a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} = c \cr
& \Rightarrow a + b - c = - 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} \cr
& {\text{Again take cube both sides}} \cr
& \Rightarrow {\left( {a + b - c} \right)^3} = - 27abc \cr
& \Rightarrow {\left( {a + b - c} \right)^3} + 27abc = 0 \cr
& \cr
& {\bf{Alternate:}} \cr
& {\text{Put value of }} \cr
& a = 0 \cr
& b = 1 \cr
& c = 1 \cr
& {\text{Value of }}{\left( {a + b - c} \right)^3} + 27abc \cr
& = {\left( {0 + 1 - 1} \right)^3} + 27 \times 0 \times 1 \times 1 \cr
& = 0 \cr} $$