1. **State the problem:** Find the only odd single-digit number greater than 1 whose cube is a perfect square. 2. **Understand the problem:** We want an odd number $n$ such that $1 < n < 10$ and $n^3$ is a perfect square. 3. **Recall important rules:** - A perfect square is a number that can be expressed as $k^2$ for some integer $k$. - For $n^3$ to be a perfect square, the prime factorization of $n$ must have exponents that make $3 \times$ those exponents even (since $n^3 = (p_1^{a_1} p_2^{a_2} \cdots)^3 = p_1^{3a_1} p_2^{3a_2} \cdots$). 4. **Check odd single-digit numbers greater than 1:** These are $3, 5, 7, 9$. 5. **Test each:** - $3^3 = 27$, prime factorization $3^3$, exponent 3 (odd), so $27$ is not a perfect square. - $5^3 = 125$, prime factorization $5^3$, exponent 3 (odd), not a perfect square. - $7^3 = 343$, prime factorization $7^3$, exponent 3 (odd), not a perfect square. - $9^3 = (3^2)^3 = 3^{6}$, exponent 6 (even), so $9^3 = 729$ is a perfect square. 6. **Conclusion:** The only odd single-digit number greater than 1 whose cube is a perfect square is $9$. **Final answer:** $9$