If $$\root 3 \of a + \root 3 \of b = \root 3 \of c {\text{,}}$$    then the simplest value of $${\left( {a + b - c} \right)^3}$$   + $$27abc$$   is?

Solution(By Examveda Team)

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