1. **State the problem:** Simplify the expression $\frac{(q r p^{-2})^{-2} p^{-2} q^{5} r^{0}}{q r}$. 2. **Recall the rules:** - Power of a product: $(ab)^n = a^n b^n$. - Power of a power: $(a^m)^n = a^{mn}$. - Negative exponents: $a^{-n} = \frac{1}{a^n}$. - Any number to the zero power is 1: $r^0 = 1$. 3. **Apply the power of a product rule:** $$(q r p^{-2})^{-2} = q^{-2} r^{-2} p^{4}$$ 4. **Rewrite the numerator:** $$q^{-2} r^{-2} p^{4} \cdot p^{-2} \cdot q^{5} \cdot r^{0} = q^{-2} r^{-2} p^{4} p^{-2} q^{5} \cdot 1$$ 5. **Combine like bases by adding exponents:** $$q^{-2 + 5} r^{-2} p^{4 - 2} = q^{3} r^{-2} p^{2}$$ 6. **Rewrite the denominator:** $$q^{1} r^{1}$$ 7. **Divide numerator by denominator by subtracting exponents:** $$\frac{q^{3} r^{-2} p^{2}}{q^{1} r^{1}} = q^{3 - 1} r^{-2 - 1} p^{2} = q^{2} r^{-3} p^{2}$$ 8. **Rewrite negative exponent as fraction:** $$q^{2} p^{2} r^{-3} = \frac{q^{2} p^{2}}{r^{3}}$$ **Final answer:** $$\frac{q^{2} p^{2}}{r^{3}}$$