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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