# Homework Help: 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?