1. Guess the cube root of 1728. 10³=1000 and 20³=2³x10³=8000 (or 20²X20=400x20=8000). Therefore, the cube root of 1728 is the 3rd power of some number between 10 and 20, and nearer to 10 rather than 20.*
2. Let the unknown number be X, so X=10+n (n=1,2,3,....,9).
3. From the standard form of polynomial, (a+b)³=a³+3a²b+3ab²+b³, we have: X³=(10+n)³=10³+3·10²·n+3·10·n²+n³ (n=1,2,3,...,9) ··· Eq.1
4. Notice that the ones digit of 1728 is 8. So the ones digit of n³ must be 8. Because, in Eq.1, each of 10³, 3·10²·n, and 3·10·n² is the multiple of 10, so any of them doesn't include the ones digit of 1728.
5. Check the ones digit of n³ by plugging each value of n into n³. If n=8, 8²=64, 4·8=32, so the ones digit is 2. The ones digit of n³ is: 1(n=1), 8(n=2), 7(n=3), 4(n=4), 5(n=5), 6(n=6), 3(n=7), 2(n=8) and 9(n=9). Only n=2 makes the ones digit of n³ up to 8. Therfore, X=10+2=12.
6. CK: 12³ (=12²·12=144·12=144·(10+2)=1440+288=1440+260+28=1700+28) =1728 CKD.
The answer: The cube root of 1728 is 12.
* If you noticed 13³=13²·13=169·13=169·10+169·3)>1690+100·3=1990>1728, you don't need to check all 'n's. All you need to do is to check n=1 and 2 for 11³ and 12³ to get 12³=1728.