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This question belongs to Arithmetic Ability Algebra
Algebra
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If $${\left( {a + \frac{1}{a}} \right)^2} = 3{\text{,}}$$    the value of $${a^3} + \frac{1}{{{a^3}}}$$   = ?

Answer & Solution
Correct Answer: Option A
$$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {\text{ }}{a^3} + \frac{1}{{{a^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr} $$
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