Question : If $(x+\frac{1}{x})^2=3$, then what is the value of $x^6 + x^{–6}$?
Option 1: 6
Option 2: 2
Option 3: –6
Option 4: –2
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Correct Answer: –2
Solution : Given: $(x+\frac{1}{x})^2=3$ We know the algebraic identities, $(x+\frac{1}{x})^2= x^2+\frac{1}{x^2}+2$ and $(x+\frac{1}{x})^3=x^6 + \frac{1}{x^{6}}+3(x^2+\frac{1}{x^2})$ $(x+\frac{1}{x})^2=3$ ⇒ $x^2+\frac{1}{x^2}+2=3$ ⇒ $x^2+\frac{1}{x^2}=3–2$ ⇒ $x^2+\frac{1}{x^2}=1$ (equation 1) Take the cube of equation (1) on both sides, $(x^2+\frac{1}{x^2})^3=1^3$ ⇒ $x^6 + \frac{1}{x^{6}}+3(x^2+\frac{1}{x^2})=1$ ⇒ $x^6 + \frac{1}{x^{6}}+3\times 1=1$ ⇒ $x^6 + \frac{1}{x^{6}}=1-3$ ⇒ $x^6 + {x^{–6}}=-2$ Hence, the correct answer is – 2.
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