1. **Problem Statement:** Find three numbers from 1 to 9 such that when multiplied together, the product is the same number repeated three times (i.e., the product equals $n \times n \times n = n^3$ for some integer $n$. 2. **Understanding the problem:** We want three numbers $a$, $b$, and $c$ (each between 1 and 9) such that: $$ a \times b \times c = n^3 $$ where $n$ is an integer from 1 to 9. 3. **Approach:** We will check cubes of numbers 1 through 9: $$ 1^3=1, 2^3=8, 3^3=27, 4^3=64, 5^3=125, 6^3=216, 7^3=343, 8^3=512, 9^3=729 $$ Since $a,b,c$ are from 1 to 9, their product must be one of these cubes. 4. **Check possible factorizations:** - For $n=1$, $1^3=1$, possible triple: $(1,1,1)$ - For $n=2$, $8=2^3$, possible triple: $(2,2,2)$ - For $n=3$, $27=3^3$, possible triple: $(3,3,3)$ - For $n=4$, $64=4^3$, possible triple: $(4,4,4)$ - For $n=5$, $125=5^3$, but 125 is greater than $9 \times 9 \times 9=729$, so possible triple: $(5,5,5)$ - For $n=6$, $216=6^3$, possible triple: $(6,6,6)$ - For $n=7$, $343=7^3$, possible triple: $(7,7,7)$ - For $n=8$, $512=8^3$, possible triple: $(8,8,8)$ - For $n=9$, $729=9^3$, possible triple: $(9,9,9)$ 5. **Conclusion:** The only way to multiply three numbers from 1 to 9 to get the same number three times is to multiply the same number three times, i.e., the triple $(n,n,n)$ where $n$ is from 1 to 9. **Final answer:** The triples are $(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6), (7,7,7), (8,8,8), (9,9,9)$.