1. **State the problem:** Simplify the expression $$\frac{(2 p m^{-1} q^{0})^{-4} \cdot 2 m^{-1} p^{3}}{2 p q^{2}}$$. 2. **Recall important rules:** - Any term raised to the zero power is 1, so $q^{0} = 1$. - When raising a product to a power, raise each factor to that power: $(abc)^n = a^n b^n c^n$. - Negative exponents mean reciprocal: $a^{-n} = \frac{1}{a^n}$. - When multiplying like bases, add exponents: $a^m \cdot a^n = a^{m+n}$. - When dividing like bases, subtract exponents: $\frac{a^m}{a^n} = a^{m-n}$. 3. **Simplify inside the numerator:** $$(2 p m^{-1} q^{0})^{-4} = 2^{-4} p^{-4} m^{4} q^{0}$$ Since $q^{0} = 1$, this becomes: $$2^{-4} p^{-4} m^{4}$$ 4. **Multiply by the other numerator term:** $$2^{-4} p^{-4} m^{4} \cdot 2 m^{-1} p^{3} = 2^{-4+1} p^{-4+3} m^{4-1} = 2^{-3} p^{-1} m^{3}$$ 5. **Write the entire fraction:** $$\frac{2^{-3} p^{-1} m^{3}}{2 p q^{2}}$$ 6. **Rewrite the denominator with exponents:** $$2^{1} p^{1} q^{2}$$ 7. **Divide numerator by denominator by subtracting exponents:** $$2^{-3-1} p^{-1-1} m^{3} q^{-2} = 2^{-4} p^{-2} m^{3} q^{-2}$$ 8. **Rewrite with positive exponents:** $$\frac{m^{3}}{2^{4} p^{2} q^{2}} = \frac{m^{3}}{16 p^{2} q^{2}}$$ **Final answer:** $$\boxed{\frac{m^{3}}{16 p^{2} q^{2}}}$$