Write 1729 as the sum of two cubes

1. Feb 28, 2017

albertrichardf

1. The problem statement, all variables and given/known data
$1729$ can be written as $12^3 + 1^3$ and $9^3 + 10^3$ and $7(10 + 9)(12 + 1)$. If
$x^3 + (7 - x)^3 = 1729$, use the above to find $x$. $x$ is a non-integer

2. Relevant equations
$1729 = 12^3 + 1^3 = 9^3 + 10^3 = 7(10 + 9)(12 + 1) = x^3 + (7 - x)^3$

3. The attempt at a solution
I used the last equation (the one with x in it) and equated it to 1729 to solve for $x$. But I need to solve the question without a calculator, and I need to use the information given in the question, which I did not.
I noticed that the 3rd form equation $1729 = 7(10 + 9)(12 + 1)$ contains the same terms as those in the cubes, $(12 + 1)$ for $(12^3 + 1^3)$ and $(10 + 9)$ for $(10^3 + 9^3)$. I think this is important, but I have no idea of how to use this information.
Thank you for helping

2. Feb 28, 2017

Khashishi

Do you remember how to factor $(a^3 + b^3)$? Try that and see where it goes.

3. Feb 28, 2017

albertrichardf

I obtain the following:

$1729 = 7(9 + 10)(12 + 1) = 7(x^2 - x(7 - x) + (7-x)^2) = (12 + 1)(12^2 - 12 + 1) = (9 + 10)(9^2 - 90 + 10^2)$

Edit: wrongly wrote a term

Last edited: Mar 1, 2017
4. Feb 28, 2017

haruspex

I think you mean (7-x)2 as the last term.

5. Mar 1, 2017

albertrichardf

Oh yes. Sorry. Thank you for correcting me

6. Mar 2, 2017

SammyS

Staff Emeritus
Now that you have made that correction, how well are you progressing toward a solution?

7. Mar 2, 2017

Staff: Mentor

Have you tried what Khashishi suggested in post #2?

8. Mar 2, 2017

haruspex

Isn't that what led to post #3?

9. Mar 3, 2017

Staff: Mentor

I thought he meant $$(a+b)(a^2-ab+b^2)$$

10. Mar 3, 2017

Staff: Mentor

I would have set this up differently.
$$(a+b)[(a+b)^2-3ab]=n[n^2-3x(n-x)]$$where n is a factor of 1729.

11. Mar 3, 2017

haruspex

What is an obvious step from there?