If $$c + \frac{1}{c} = \sqrt 3 {\text{,}}$$ then the value of $${c^3} + \frac{1}{{{c^3}}}$$ is equal to? Options: A . &nbsp0 B . &nbsp$$\frac{3}{{\sqrt 3 }}$$ C . &nbsp$$\frac{1}{{\sqrt 3 }}$$ D . &nbsp$$\frac{6}{{\sqrt 3 }}$$

If $$c + \frac{1}{c} = \sqrt 3 {\text{,}}$$ then the value of $${c^3} + \frac

Question

If $$c + \frac{1}{c} = \sqrt 3 {\text{,}}$$ then the value of $${c^3} + \frac{1}{{{c^3}}}$$ is equal to?

Options:

A . 0

B . $$\frac{3}{{\sqrt 3 }}$$

C . $$\frac{1}{{\sqrt 3 }}$$

D . $$\frac{6}{{\sqrt 3 }}$$

Answer: Option A
$$\eqalign{
& c + \frac{1}{c} = \sqrt 3 \cr
& {\text{On cubing both side}} \cr
& \Rightarrow {\left( {c + \frac{1}{c}} \right)^3} = 3\sqrt 3 \cr
& \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3.c.\frac{1}{c}\left( {c + \frac{1}{c}} \right) = 3\sqrt 3 \cr
& \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr
& \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr
& \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 0 \cr} $$

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