If $${a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} + {c^{\frac{1}{3}}} = 0,$$ then a relation among a, b, c is?
A. a + b + c = 0
B. (a + b + c)3 = 27abc
C. a + b + c = 3abc
D. a3 + b3 + c3 = 0
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} + {c^{\frac{1}{3}}} = 0 \cr & \Rightarrow {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} = - {c^{\frac{1}{3}}} \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right)^3} = {\left( { - {c^{\frac{1}{3}}}} \right)^3} \cr & \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right) = - c \cr & \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }} - {c^{\frac{1}{3}}}} \right) = - c \cr & \Rightarrow a + b + c = 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} \cr & {\text{Again taking cube }} \cr & \Rightarrow {\left( {a + b + c} \right)^3} = 27abc \cr} $$This Question Belongs to Arithmetic Ability >> Algebra
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