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