Question

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

Answer 1

$x=2+\sqrt{3}$

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

(Rationalising the denominator)

$=\frac{2-\sqrt{3}}{(2{)}^2-(\sqrt{3}{)}^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$

$\therefore x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4$

Cubing both sides,

${(x+\frac{1}{x})}^3=(4{)}^3$

$\Rightarrow x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=64$

$\Rightarrow x^3+\frac{1}{x^3}+3\times 4=64$

$\Rightarrow x^3+\frac{1}{x^3}+12=64$

$\Rightarrow x^3+\frac{1}{x^3}=64-12=52$

Hence $x^3+\frac{1}{x^3}=52$