Question

If $x=\left(2+\sqrt{3}\right)$, show that $\left(x^3+\frac{1}{x^3}\right)=52$.

Answer 1

$$\begin{gathered} x=2+\sqrt{3} \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ \frac{1}{x}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}} \\ =\frac{2-\sqrt{3}}{4-3} \\ =2-\sqrt{3} \\ \mathrm{Now}, \\ x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4 \\ \Rightarrow {\left(x+\frac{1}{x}\right)}^3=4^3 \\ \Rightarrow x^3+{\left({\displaystyle \frac{1}{x}}\right)}^3+3\times x\times \frac{1}{x}\times \left(x+\frac{1}{x}\right)=64 \\ \Rightarrow x^3+{\left({\displaystyle \frac{1}{x}}\right)}^3+3\times \left(4\right)=64\left(\because x+\frac{1}{x}=4\right) \\ \Rightarrow x^3+{\left(\frac{1}{x}\right)}^3+12=64 \\ \Rightarrow x^3+{\left(\frac{1}{x}\right)}^3=64-12 \\ \Rightarrow x^3+{\left(\frac{1}{x}\right)}^3=52 \end{gathered}$$