1. The problem is to simplify the expression $$\left(\frac{2}{3}\right)^{-3}$$ and verify that it equals $$\left(\frac{3}{2}\right)^3$$ and $$\frac{27}{8}$$. 2. Start by applying the negative exponent rule: $$a^{-n} = \frac{1}{a^n}$$. 3. So, $$\left(\frac{2}{3}\right)^{-3} = \frac{1}{\left(\frac{2}{3}\right)^3}$$. 4. Calculate the cube: $$\left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27}$$. 5. Substitute back: $$\frac{1}{\frac{8}{27}} = \frac{27}{8}$$. 6. Also check $$\left(\frac{3}{2}\right)^3 = \frac{3^3}{2^3} = \frac{27}{8}$$. 7. Therefore, both expressions are equal to $$\frac{27}{8}$$, confirming the simplification. Final answer: $$\left(\frac{2}{3}\right)^{-3} = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$$.