Question : If $x^2-\sqrt{7} x+1=0$, then $\left(x^3+x^{-3}\right)=?$
Option 1: $4 \sqrt{7}$
Option 2: $3 \sqrt{7}$
Option 3: $10 \sqrt{7}$
Option 4: $7 \sqrt{7}$
Correct Answer: $4 \sqrt{7}$
Solution :
Given: $x^2-\sqrt{7} x+1=0$
Dividing both sides by $x$, we get:
⇒ $x-\sqrt{7}+\frac{1}{x}=0$
⇒ $x+\frac{1}{x}=\sqrt{7}$
Cubing both sides, we get:
⇒ $(x+\frac{1}{x})^3=(\sqrt{7})^3$
⇒ $x^3+\frac{1}{x^3}+3(x\times\frac{1}{x})(x+\frac{1}{x})=(\sqrt{7})^3$
⇒ $x^3+\frac{1}{x^3}+3\times\sqrt7=7\sqrt{7}$
⇒ $x^3+\frac{1}{x^3}=4\sqrt{7}$
Hence, the correct answer is $4\sqrt{7}$.
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