Question

Proof that `N = (45+29 sqrt(2))^(1/3)+(45-29 sqrt(2))^(1/3)` is a integer ?

Answer 1

Consider `t^3-21t-90 = 0`

This has one Real root which is `6` a.k.a. `(45+29sqrt(2))^(1/3)+(45-29sqrt(2))^(1/3)`

Explanation:

Consider the equation:

`t^3-21t-90 = 0`

Using Cardano's method to solve it, let `t = u+v`

Then:

`u^3+v^3+3(uv-7)(u+v)-90 = 0`

To eliminate the term in `(u+v)`, add the constraint `uv=7`

Then:

`u^3+7^3/u^3-90 = 0`

Multiply through by `u^3` and rearrange to get the quadratic in `u^3`:

`(u^3)^2-90(u^3)+343 = 0`

by the quadratic formula, this has roots:

`u^3 = (90+-sqrt(90^2-(4*343)))/2`

`color(white)(u^3) = 45 +- 1/2sqrt(8100-1372)`

`color(white)(u^3) = 45 +- 1/2sqrt(6728)`

`color(white)(u^3) = 45 +- 29sqrt(2)`

Since this is Real and the derivation was symmetric in `u` and `v`, we can use one of these roots for `u^3` and the other for `v^3` to deduce that the Real zero of `t^3-21t-90` is:

`t_1 = root(3)(45+29sqrt(2))+root(3)(45-29sqrt(2))`

but we find:

`(6)^3-21(6)-90 = 216 - 126 - 90 = 0`

So the Real zero of `t^3-21t-90` is `6`

So `6 = root(3)(45+29sqrt(2))+root(3)(45-29sqrt(2))`

`color(white)()`
Footnote

To find the cubic equation, I used Cardano's method backwards.

Answer 2

`N = 6`

Explanation:

Making `x = 45+29 sqrt(2)` and `y = 45-29 sqrt(2)` then

`(x^(1/3)+y^(1/3))^3=x + 3 (x y)^(1/3)x^(1/3)+3(x y)^(1/3) y^(1/3)+y`

`(x y)^(1/3) = (7^3)^(1/3) = 7`
`x+y =2 xx 45`

so

`(x^(1/3)+y^(1/3))^3 = 90 + 21(x^(1/3)+y^(1/3))`

or calling `z = x^(1/3)+y^(1/3)` we have

`z^3-21 z-90 = 0`

with `90 = 2 xx 3^2 xx 5` and `z = 6` is a root so

`x^(1/3)+y^(1/3) = 6`