Question : If $\left(x-\frac{1}{x}\right) =4$, then what is the value of $\left(x^6+\frac{1}{x^6}\right)$?
Option 1: 4689
Option 2: 4786
Option 3: 5832
Option 4: 5778
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Correct Answer: 5778
Solution : Given: $x-\frac{1}{x}=4$ Squaring both sides, ⇒ $x^2+\frac{1}{x^2} - 2 = 16$ ⇒ $x^2+\frac{1}{x^2} = 18$ Now cubing both sides, $(x^2+\frac{1}{x^2})^3 = 18^3$ ⇒ $x^6+\frac{1}{x^6}+3×x^2×\frac{1}{x^2}(x^2+\frac{1}{x^2}) = 5832$ ⇒ $x^6+\frac{1}{x^6}+3(18) = 5832$ ⇒ $x^6+\frac{1}{x^6} = 5832-54 = 5778$ Hence, the correct answer is 5778.
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