Question

If $x=2+\sqrt{3}$, then $x^3+\frac{1}{x^3}$ = __.

• A. 64
• B. 52
• C. 48
• D. 4

Answer 1

The correct option is B

52

$x=2+\sqrt{3}$ ...(i)

$\frac{1}{x}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$

=$\frac{2-\sqrt{3}}{(2{)}^2-(\sqrt{3}{)}^2}$

=$\frac{2-\sqrt{3}}{4-3}$

=$2-\sqrt{3}$ ...(ii)

$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}$

= $4$

$\therefore x^3+\frac{1}{x^3}=(x+\frac{1}{x}{)}^3-3(x+\frac{1}{x})$

=$(4{)}^3-3(4)$

=$64-12$

=$52$