1. **State the problem:** Simplify the expression $$- y^{-4} z^{2} \cdot y^{0} z^{2} \div (2z^{-4})^{3}$$. 2. **Recall the rules:** - Any number or variable to the zero power is 1, so $y^{0} = 1$. - When multiplying powers with the same base, add exponents: $a^{m} \cdot a^{n} = a^{m+n}$. - When dividing powers with the same base, subtract exponents: $\frac{a^{m}}{a^{n}} = a^{m-n}$. - Power of a power: $(a^{m})^{n} = a^{m \cdot n}$. - Negative exponents mean reciprocal: $a^{-m} = \frac{1}{a^{m}}$. 3. **Simplify numerator:** $$- y^{-4} z^{2} \cdot y^{0} z^{2} = - y^{-4 + 0} z^{2 + 2} = - y^{-4} z^{4}$$ 4. **Simplify denominator:** $$(2 z^{-4})^{3} = 2^{3} (z^{-4})^{3} = 8 z^{-12}$$ 5. **Rewrite the expression:** $$\frac{- y^{-4} z^{4}}{8 z^{-12}}$$ 6. **Divide powers of $z$:** $$z^{4} \div z^{-12} = z^{4 - (-12)} = z^{16}$$ 7. **Combine all:** $$\frac{- y^{-4} z^{16}}{8} = - \frac{z^{16}}{8 y^{4}}$$ **Final answer:** $$- \frac{z^{16}}{8 y^{4}}$$