Rationalization of a fraction

Question

Can someone please explain how i can rationalize the fraction $\frac{2 +\sqrt{3}}{(2-\sqrt{3})^3}$ so that i obtain an answer that has an exponent of 6. I basically need to compare this value with $\dfrac{(10)^6}{(3)^6}$ to see which is greater. Thanks in advance for helping out!

Answer 1

Hint : Multiply numerator and denominator with $$(2+\sqrt{3})^3$$

Answer 2

\begin{align*} \frac{2+\sqrt{3}}{(2-\sqrt{3})^3}&=\frac{2+\sqrt{3}}{(2-\sqrt{3})^3}\cdot\frac{(2+\sqrt{3})^3}{(2+\sqrt{3})^3}\\ &=\frac{(2+\sqrt{3})^4}{(2^2-3)^3}\\ &=\frac{(2+\sqrt{3})^4}{1}\\ &=(2+\sqrt{3})^4 \end{align*}

Answer 3

You could multiply top and bottom of your fraction by $(2+\sqrt{3})^3$ and obtain $(2+\sqrt{3})^4$ since $(2+\sqrt{3})(2-\sqrt{3})=1$