Answer: Option A
$$\eqalign{
& \frac{1}{{\root 3 \of 4 + \root 3 \of 2 + 1}} = a\root 3 \of 4 + b\root 3 \of 2 + c \cr
& \Rightarrow \frac{1}{{\root 3 \of 4 + \root 3 \of 2 + 1}} = \frac{1}{{{{\left( {{2^{\frac{1}{3}}}} \right)}^2} + {2^{\frac{1}{3}}} + {{\left( 1 \right)}^2}}} \cr
& \Rightarrow \therefore {{\text{A}}^3} - {{\text{B}}^3}{\text{ = }}\left( {{\text{A}} - {\text{B}}} \right)\left( {{{\text{A}}^2} + {\text{AB}} + {{\text{B}}^2}} \right) \cr
& {\text{Put, A}} = {2^{\frac{1}{3}}}{\text{, B}} = 1 \cr
& \Rightarrow \frac{{\left( {{2^{\frac{1}{3}}} - 1} \right)}}{{\left( {{2^{\frac{1}{3}}} - 1} \right)\left( {{{\left( {{2^{\frac{1}{3}}}} \right)}^2} + {2^{\frac{1}{3}}} + {{\left( 1 \right)}^2}} \right)}} \cr
& \Rightarrow \frac{{\left( {{2^{\frac{1}{3}}} - 1} \right)}}{{{{\left( {{2^{\frac{1}{3}}}} \right)}^3} - {{\left( 1 \right)}^3}}} \cr
& \Rightarrow \left( {{2^{\frac{1}{3}}} - 1} \right) \cr
& \therefore {2^{\frac{1}{3}}} - 1 \cr
& = a\left( {{2^{\frac{2}{3}}}} \right) + b{\left( 2 \right)^{\frac{1}{3}}} + c \cr
& \left( {{\text{Comparing the terms}}} \right) \cr
& a = 0 \cr
& b = 1 \cr
& c = - 1 \cr
& \therefore a + b + c \cr
& = 0 + 1 - 1 \cr
& = 0 \cr} $$