1. **State the problem:** Simplify the expression $$\frac{(2^{-3} \cdot 2^{10})^{-2}}{2^{3} (2^{-5} \cdot 2^{-2})^{3}}$$. 2. **Recall the laws of exponents:** - Product of powers: $$a^{m} \cdot a^{n} = a^{m+n}$$ - Power of a power: $$(a^{m})^{n} = a^{m \cdot n}$$ - Quotient of powers: $$\frac{a^{m}}{a^{n}} = a^{m-n}$$ 3. **Simplify inside the parentheses:** - Numerator inside parentheses: $$2^{-3} \cdot 2^{10} = 2^{-3+10} = 2^{7}$$ - Denominator inside parentheses: $$2^{-5} \cdot 2^{-2} = 2^{-5-2} = 2^{-7}$$ 4. **Apply the outer exponents:** - Numerator: $$(2^{7})^{-2} = 2^{7 \cdot (-2)} = 2^{-14}$$ - Denominator second term: $$(2^{-7})^{3} = 2^{-7 \cdot 3} = 2^{-21}$$ 5. **Rewrite the entire expression:** $$\frac{2^{-14}}{2^{3} \cdot 2^{-21}}$$ 6. **Simplify the denominator:** $$2^{3} \cdot 2^{-21} = 2^{3 + (-21)} = 2^{-18}$$ 7. **Now the expression is:** $$\frac{2^{-14}}{2^{-18}}$$ 8. **Apply quotient of powers rule:** $$2^{-14 - (-18)} = 2^{-14 + 18} = 2^{4}$$ 9. **Final answer:** $$\boxed{16}$$