1. **State the problem:** Given the equation $$(a^b)^n = \frac{256}{729}$$, find the values of $a$, $b$, and $n$. 2. **Rewrite the equation using exponent rules:** Recall that $$(a^b)^n = a^{bn}$$. 3. **Express the right side as powers of prime bases:** $$256 = 2^8$$ $$729 = 3^6$$ So, $$\frac{256}{729} = \frac{2^8}{3^6} = 2^8 \times 3^{-6}$$ 4. **Assume $a$ is a fraction with bases 2 and 3:** Let $$a = \frac{2^m}{3^k}$$ for some integers $m$ and $k$. 5. **Apply the exponent:** $$a^{bn} = \left(\frac{2^m}{3^k}\right)^{bn} = 2^{mbn} \times 3^{-kbn}$$ 6. **Equate exponents from both sides:** $$mbn = 8$$ $$-kbn = -6 \implies kbn = 6$$ 7. **From these, we have two equations:** $$mbn = 8$$ $$kbn = 6$$ 8. **Divide the first equation by the second:** $$\frac{mbn}{kbn} = \frac{8}{6} \implies \frac{m}{k} = \frac{4}{3}$$ 9. **Choose simplest integer values for $m$ and $k$ satisfying this ratio:** $$m = 4, \quad k = 3$$ 10. **Use $mbn = 8$ to find $bn$:** $$4bn = 8 \implies bn = 2$$ 11. **Use $kbn = 6$ to find $bn$:** $$3bn = 6 \implies bn = 2$$ 12. **Since $bn = 2$, $b$ and $n$ can be any pair of numbers whose product is 2.** **Final answer:** $$a = \frac{2^4}{3^3} = \frac{16}{27}, \quad b, n \text{ such that } bn = 2$$