Question

What is the root of `x^3 - x - 1 = 0` in exact form?

Answer 1

Real root:

`root(3)((27+3sqrt(69))/54)+root(3)((27-3sqrt(69))/54)`

and related complex roots...

Explanation:

Given:

`x^3-x-1 = 0`

Using Cardano's method, let `x = u+v`

Then:

`u^3+v^3+(3uv-1)(u+v)-1 = 0`

To eliminate the term in `(u+v)` add the constraint:

`3uv-1 = 0`

So:

`v = 1/(3u)`

and our equation can be rewritten:

`u^3+1/(27u^3)-1 = 0`

Multiply through by `27u^3` and rearrange a little to get:

`27(u^3)^2-27(u^3)+1 = 0`

Hence by the quadratic formula:

`u^3 = (27+-sqrt(27^2-4(27)(1)))/(2*27)`

`color(white)(u^3) = (27+-sqrt(729-108))/54`

`color(white)(u^3) = (27+-sqrt(621))/54`

`color(white)(u^3) = (27+-3sqrt(69))/54`

Since these are both real values and the derivation is symmetric in `u` and `v`, we can use one for `u^3` and the other for `v^3` to find real root of our original cubic:

`x_1 = root(3)((27+3sqrt(69))/54)+root(3)((27-3sqrt(69))/54)`

and related complex roots:

`x_2 = omega root(3)((27+3sqrt(69))/54)+omega^2 root(3)((27-3sqrt(69))/54)`

`x_3 = omega^2 root(3)((27+3sqrt(69))/54)+omega root(3)((27-3sqrt(69))/54)`

where `omega = -1/2+sqrt(3)/2i` is the primitive complex cube root of `1`.