Basic problem: express 1729 as the sum of 2 cubes

[Bladesofhalo]

I have to express 1729 as the sum of 2 cubes.

I know there are 2 ways to do it, but could someone show me how to do it one way and I will figure out the rest. Thanks.

 

[galactus]

How about \(\displaystyle 10^{3}+9^{3}\)

Factor using sum of two cubes:

\(\displaystyle (10+9)(10^{2}-(10)(9)+9^{2})\)

 

[Denis]

Hint (looking last digit):
0^3 = 0, 9^3 = 9 : 0 + 9 = 9
1^3 = 1, 2^3 = 8 : 1 + 8 = 9
3^3 = 7, 8^3 = 2 : 7 + 2 = 9
4^3 = 4, 5^3 = 5 : 4 + 5 = 9
6^6 = 6, 7^3 = 3 : 6 + 3 = 9

That'll lead you to 2 solutions: 1,12 and 9,10

 

[soroban]

Hello, Bladesofhalo!

I have to express 1729 as the sum of 2 cubes.
I know there are 2 ways to do it.

I know of no elementary method for finding them.

You might as well make a list of cubes and try to get a sum of 1729.

. . . \(\displaystyle \begin{array}{cccccccccccc}1^3 & = & 1 \\ 2^3 & = & 8 \\ 3^3 & = & 27 \\ 4^3 & = & 64 \\ 5^3 & = & 125 \\ 6^3 & = & 216 \\ 7^3 & = & 343 \\ 8^3 & = & 512 \\ 9^3 & = & 729 \\ 8^3 & = & 512 \\ 9^3 & = & 729 \\ 10^3 & = & 1000 \\ 11^3 & = & 1331 \\ 12^3 & = & 1728\end{array}\)