Question

If ${(x+\frac{1}{x})}^2=3$, then find the value of $x^3+\frac{1}{x^3}$.

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A 0

${(x+\frac{1}{x})}^2=3\Rightarrow x+\frac{1}{x}=\pm \sqrt{3}$
Now ${(x+\frac{1}{x})}^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})$
$\Rightarrow x^3+\frac{1}{x^3}={(x+\frac{1}{x})}^3-3(x+\frac{1}{x})=(\pm \sqrt{3}{)}^3-3\times (\pm \sqrt{3})$
$=\pm 3\sqrt{3}\pm 3\sqrt{3}=(3\sqrt{3}-3\sqrt{3})or(-3\sqrt{3}+3\sqrt{3})=0$