Write 1729 as the sum of two cubes

[albertrichardf]

Homework Statement

$1729$ 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

Homework Equations

$1729 = 12^3 + 1^3 = 9^3 + 10^3 = 7(10 + 9)(12 + 1) = x^3 + (7 - x)^3$

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

 

[Khashishi]

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

 

[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

 

[haruspex]

7(x²−x(7−x)+7²)

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

 

[albertrichardf]

Oh yes. Sorry. Thank you for correcting me

 

[SammyS]

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

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

 

[Chestermiller]

Have you tried what Khashishi suggested in post #2?

 

[haruspex]

Have you tried what Khashishi suggested in post #2?

Isn't that what led to post #3?

 

[Chestermiller]

Isn't that what led to post #3?

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

 

[Chestermiller]

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.

 

[haruspex]

7(9+10)(12+1)=7(x²−x(7−x)+(7−x)²)

What is an obvious step from there?