Show that :

$x^3 + \dfrac{1}{x^3} = 52$, if x = 2 + $\sqrt{3}$

(i) Given,

x = 2 + $\sqrt{3}$

$\therefore \dfrac{1}{x} = \dfrac{1}{2 + \sqrt{3}}$

Rationalizing,

$\Rightarrow \dfrac{1}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} \\[1em] \Rightarrow \dfrac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \dfrac{2 - \sqrt{3}}{4 - 3} \\[1em] \Rightarrow 2 - \sqrt{3}. \\[1em] \therefore \dfrac{1}{x} = 2 - \sqrt{3}$

Substituting value of x and $\dfrac{1}{x} \text{ in } x^3 + \dfrac{1}{x^3}$, we get :

$\Rightarrow x^3 + \dfrac{1}{x^3} = (2 + \sqrt{3})^3 + (2 -\sqrt{3})^3 \\[1em] = 2^3 + (\sqrt{3})^3 + 3 \times 2 \times \sqrt{3} \times (2 + \sqrt{3}) + 2^3 - (\sqrt{3})^3 - 3 \times 2 \times \sqrt{3} \times (2 - \sqrt{3}) \\[1em] = 8 + 3\sqrt{3} + 6\sqrt{3}(2 + \sqrt{3}) + 8 - 3\sqrt{3} - 6\sqrt{3}(2 - \sqrt{3}) \\[1em] = 8 + 8 + 3\sqrt{3} - 3\sqrt{3} + 12\sqrt{3} + 18 - 12\sqrt{3} + 18 \\[1em] = 8 + 8 + 18 + 18 \\[1em] = 52.$

Hence, proved that $x^3 + \dfrac{1}{x^3} = 52$.