Question

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

Answer 1

$$\begin{gathered} x=2+\sqrt{3}.....\left(1\right) \\ \Rightarrow \frac{1}{x}=\frac{1}{2+\sqrt{3}} \\ \Rightarrow \frac{1}{x}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{1}{x}=\frac{2-\sqrt{3}}{2^2-{\left(\sqrt{3}\right)}^2} \\ \Rightarrow \frac{1}{x}=\frac{2-\sqrt{3}}{4-3} \\ \Rightarrow \frac{1}{x}=2-\sqrt{3}.....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4.....\left(3\right)$
Cubing both sides, we get
$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^3=4^3 \\ \Rightarrow x^3+\frac{1}{x^3}+3\times x\times \frac{1}{x}\left(x+\frac{1}{x}\right)=64 \end{gathered}$$
$\Rightarrow x^3+\frac{1}{x^3}+3\times 4=64$ [Using (3)]
$\Rightarrow x^3+\frac{1}{x^3}=64-12=52$
Thus, the value of $x^3+\frac{1}{x^3}$ is 52.