Jason
Let's start with another law of exponents.
\(\displaystyle integer\ q > 0 \implies p^{1/q} = \sqrt[q]{p}\ BY\ DEFINITION.\)
So \(\displaystyle integer\ q >0 \implies p^{r/q} = \left(p^r\right)^{1/q} = \sqrt[q]{p^r}.\)
Therefore
\(\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{1}{\left(\dfrac{3}{2}\right)^{(1/3)}}.\) Negative exponent law. OK so far?
\(\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{1}{\sqrt[3]{\dfrac{3}{2}}}.\) Fractional exponent law. Still with me?
\(\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{1}{\sqrt[3]{1.5}}.\) Simplification.
Actually it is easier to simplify as the first step but I wanted to concentrate on the laws of exponents.
EDIT: Pappua and I arrive at the same answer, just expressed differently
\(\displaystyle \left(\dfrac{1}{2} * \sqrt[3]{18}\right)^3 = \dfrac{1}{27} * 18 = \dfrac{2}{3}.\)
\(\displaystyle \left(\dfrac{1}{\sqrt[3]{1.5}}\right)^3 = \dfrac{1}{1.5} = \dfrac{1}{1.5} * \dfrac{2}{2} = \dfrac{2}{3}.\)
He used a different law of exponents.
\(\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{3^{-(1/3)}}{2^{-(1/3)}}.\) He started here.
So \(\displaystyle \left(\dfrac{3}{2}\right)^{-(1/3)} = \dfrac{\dfrac{1}{3^{(1/3)}}}{\dfrac{1}{2^{(1/3)}}} = \dfrac{1}{\sqrt[3]{3}} * \dfrac{\sqrt[3]{2}}{1} = \dfrac{\sqrt[3]{2}}{\sqrt[3]{3}} =\dfrac{2^{(1/3)}}{3^{(1/3)}} = \left(\dfrac{2}{3}\right)^{(1/3)}.\) Now he rationalized.
\(\displaystyle \left(\dfrac{2}{3}\right)^{-(1/3)} = \left(\dfrac{2}{3}\right)^{(1/3)} = \left(\dfrac{2 * 9}{3 * 9}\right)^{(1/3)} = \left(\dfrac{18}{27}\right)^{(1/3)} = \dfrac{18^{(1/3)}}{27^{(1/3)}} = \dfrac{\sqrt[3]{18}}{3}.\)
