1. **State the problem:** Evaluate the expression $$\frac{x^2}{y^{-3}} = \frac{y^3}{x^2}$$ when $$x=3$$ and $$y=-4$$. 2. **Recall the rules:** - Negative exponents mean reciprocal: $$a^{-n} = \frac{1}{a^n}$$. - Powers apply to both positive and negative numbers. 3. **Rewrite the left side:** $$\frac{x^2}{y^{-3}} = x^2 \times y^3$$ because dividing by $$y^{-3}$$ is multiplying by $$y^3$$. 4. **Substitute values:** $$x^2 \times y^3 = 3^2 \times (-4)^3$$. 5. **Calculate powers:** $$3^2 = 9$$. $$(-4)^3 = -64$$. 6. **Multiply:** $$9 \times (-64) = -576$$. 7. **Check the right side:** $$\frac{y^3}{x^2} = \frac{(-4)^3}{3^2} = \frac{-64}{9}$$. 8. **Compare both sides:** Left side = $$-576$$, Right side = $$-\frac{64}{9}$$. They are not equal numerically, so the equation is not true for these values. **Final answer:** $$\frac{x^2}{y^{-3}} = -576$$ when $$x=3$$ and $$y=-4$$.