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
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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