1. **State the problem:** Simplify the expression $$\frac{27}{(-2)^5} \cdot \frac{4^8}{6^4}$$. 2. **Recall the rules:** - Powers of negative numbers: $$(-a)^n = (-1)^n \cdot a^n$$. - Multiply fractions by multiplying numerators and denominators. - Simplify powers and factor where possible. 3. **Calculate powers:** - $$(-2)^5 = (-1)^5 \cdot 2^5 = -1 \cdot 32 = -32$$ - $$4^8 = (2^2)^8 = 2^{16}$$ - $$6^4 = (2 \cdot 3)^4 = 2^4 \cdot 3^4 = 16 \cdot 81 = 1296$$ 4. **Rewrite the expression:** $$\frac{27}{-32} \cdot \frac{2^{16}}{1296} = \frac{27 \cdot 2^{16}}{-32 \cdot 1296}$$ 5. **Factor numerator and denominator:** - 27 = $3^3$ - 32 = $2^5$ - 1296 = $2^4 \cdot 3^4$ So, $$\frac{3^3 \cdot 2^{16}}{-2^5 \cdot 2^4 \cdot 3^4} = \frac{3^3 \cdot 2^{16}}{-2^{9} \cdot 3^4}$$ 6. **Cancel common factors:** $$\frac{\cancel{3^3} \cdot 2^{16}}{-2^{9} \cdot \cancel{3^3} \cdot 3} = \frac{2^{16}}{-2^{9} \cdot 3}$$ 7. **Simplify powers of 2:** $$\frac{2^{16}}{2^{9}} = 2^{16-9} = 2^7 = 128$$ 8. **Final simplified expression:** $$\frac{128}{-3} = -\frac{128}{3}$$ **Answer:** $$-\frac{128}{3}$$