1. **State the problem:** Simplify the expression $$\frac{7^2 \times (-8)^{-5} \times 3^4}{7^5 \times 8^{-7} \times 3^3}$$. 2. **Recall the rules:** - When dividing powers with the same base, subtract the exponents: $$\frac{a^m}{a^n} = a^{m-n}$$. - Negative exponents mean reciprocal: $$a^{-n} = \frac{1}{a^n}$$. - Multiplication and division are performed separately for each base. 3. **Apply the rules to each base:** - For base 7: $$\frac{7^2}{7^5} = 7^{2-5} = 7^{-3}$$ - For base 8 (note the negative signs): $$\frac{(-8)^{-5}}{8^{-7}} = (-8)^{-5} \times 8^{7}$$ Since $$(-8)^{-5} = \frac{1}{(-8)^5}$$ and $$8^7$$ is positive, rewrite as: $$\frac{1}{(-8)^5} \times 8^7 = \frac{8^7}{(-8)^5}$$ 4. **Simplify the 8 terms:** Express $$(-8)^5 = (-1)^5 \times 8^5 = -8^5$$, so: $$\frac{8^7}{-8^5} = -\frac{8^7}{8^5} = -8^{7-5} = -8^2$$ 5. **For base 3:** $$\frac{3^4}{3^3} = 3^{4-3} = 3^1 = 3$$ 6. **Combine all simplified parts:** $$7^{-3} \times (-8^2) \times 3 = -7^{-3} \times 8^2 \times 3$$ 7. **Rewrite negative exponent:** $$7^{-3} = \frac{1}{7^3}$$, so: $$-\frac{8^2 \times 3}{7^3}$$ 8. **Calculate powers:** $$8^2 = 64$$ $$7^3 = 343$$ 9. **Final simplified expression:** $$-\frac{64 \times 3}{343} = -\frac{192}{343}$$ **Answer:** $$-\frac{192}{343}$$