Question
If $$x + \frac{1}{x} = \sqrt 3 {\text{,}}$$ then find the value of $${x^3} + \frac{1}{{{x^3}}}$$ = ?
Answer: Option B
$$\eqalign{
& x + \frac{1}{x} = \sqrt 3 \cr
& {\text{Cubing both side}} \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3.x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 3\sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr} $$
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$$\eqalign{
& x + \frac{1}{x} = \sqrt 3 \cr
& {\text{Cubing both side}} \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3.x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 3\sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr} $$
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