1. **State the problem:** Simplify the expression $$(16^{8}y^{-8}m^{12})(2x^{4}y^{5}m^{8})$$. 2. **Recall the rules:** - When multiplying powers with the same base, add the exponents: $$a^{m} \times a^{n} = a^{m+n}$$. - When multiplying coefficients, multiply them directly. - Negative exponents mean reciprocal: $$a^{-n} = \frac{1}{a^{n}}$$. 3. **Multiply the coefficients:** $$16^{8} \times 2$$ 4. **Simplify the variables by adding exponents:** - For $y$: $$y^{-8} \times y^{5} = y^{-8+5} = y^{-3}$$ - For $m$: $$m^{12} \times m^{8} = m^{12+8} = m^{20}$$ - For $x$: only in second term, so $$x^{4}$$ remains. 5. **Rewrite the expression:** $$2 \times 16^{8} \times x^{4} \times y^{-3} \times m^{20}$$ 6. **Calculate $16^{8}$:** Since $16 = 2^{4}$, then $$16^{8} = (2^{4})^{8} = 2^{4 \times 8} = 2^{32}$$ 7. **Final simplified expression:** $$2 \times 2^{32} \times x^{4} \times y^{-3} \times m^{20} = 2^{1+32} x^{4} y^{-3} m^{20} = 2^{33} x^{4} y^{-3} m^{20}$$ 8. **Express negative exponent as fraction:** $$y^{-3} = \frac{1}{y^{3}}$$ 9. **Final answer:** $$\boxed{\frac{2^{33} x^{4} m^{20}}{y^{3}}}$$