Examveda

The value of $$\left( {{x^{\frac{1}{3}}} + {x^{ - \frac{1}{3}}}} \right)\left( {{x^{\frac{2}{3}}} - 1 + {x^{ - \frac{2}{3}}}} \right){\text{is:}}$$

A. $${x^1} + {x^{\frac{2}{3}}}$$

B. $$x + {x^{ - \frac{1}{3}}}$$

C. $${x^{\frac{1}{3}}} + {x^{ - 1}}$$

D. $$x + {x^{ - 1}}$$

Answer: Option D

Solution (By Examveda Team)

$$\eqalign{ & \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) = {a^3} + {b^3} \cr & \therefore \left( {{x^{\frac{1}{3}}} + {x^{ - \frac{1}{3}}}} \right)\left( {{x^{\frac{2}{3}}} - 1 + {x^{ - \frac{2}{3}}}} \right) \cr & = \left( {{x^{\frac{1}{3}}} + {x^{ - \frac{1}{3}}}} \right)\left( {{{\left( {{x^{\frac{1}{3}}}} \right)}^2} - {x^{ - \frac{1}{3}}}.{x^{\frac{1}{3}}} + {{\left( {{x^{ - \frac{1}{3}}}} \right)}^2}} \right) \cr & = {\left( {{x^{\frac{1}{3}}}} \right)^3} + {\left( {{x^{ - \frac{1}{3}}}} \right)^3} \cr & = x + {x^{ - 1}} \cr & = x + \frac{1}{x} \cr} $$

This Question Belongs to Arithmetic Ability >> Surds And Indices

Join The Discussion

Related Questions on Surds and Indices