Question
If $$x + \frac{1}{x} \ne 0$$ Â and $${x^3} + \frac{1}{{{x^3}}} = 0{\text{,}}$$ Â then the value $${\left( {x + \frac{1}{x}} \right)^4}$$ Â is?
Answer: Option A
$$\eqalign{
& {x^3} + \frac{1}{{{x^3}}} = 0{\text{ }} \cr
& {\text{It is possible only when}} \cr
& x + \frac{1}{x} = \sqrt 3 {\text{ }} \cr
& {\text{So, }}{\left( {x + \frac{1}{x}} \right)^4} = {\left( {\sqrt 3 } \right)^4} \cr
& \Leftrightarrow {\left( {x + \frac{1}{x}} \right)^4} = 9 \cr} $$
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$$\eqalign{
& {x^3} + \frac{1}{{{x^3}}} = 0{\text{ }} \cr
& {\text{It is possible only when}} \cr
& x + \frac{1}{x} = \sqrt 3 {\text{ }} \cr
& {\text{So, }}{\left( {x + \frac{1}{x}} \right)^4} = {\left( {\sqrt 3 } \right)^4} \cr
& \Leftrightarrow {\left( {x + \frac{1}{x}} \right)^4} = 9 \cr} $$
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