Question

How do you solve `x^3-x=1`?

Answer 1

See below.

Explanation:

For my approach, I will be using a graphical interpretation.

You can rewrite the equation as `x^3 - x - 1 = 0` as the first step.

Then graph the following: `f(x) = x^3 - x - 1`.

graph{x^3 - x - 1 [-10, 10, -5, 5]}

Click on where the graph intersects the x-axis. This point should be (1.325, 0). Therefore, the answer is `x = 1.325`.

Answer 2

Use Cardano's method to find real root:

`x_1 = 1/3(root(3)((27+3sqrt(69))/2)+root(3)((27-3sqrt(69))/2))`

`color(white)(x_1) ~~ 1.324717957`

and related complex roots.

Explanation:

Given:

`x^3-x=1`

Subtract `1` from both sides to get:

`x^3-x-1 = 0`

`color(white)()`
Discriminant

The discriminant `Delta` of a cubic polynomial in the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

In our example, `a=1`, `b=0`, `c=-1` and `d=-1`, so we find:

`Delta = 0+4+0-27+0 = -23`

Since `Delta < 0` this cubic has `1` Real zero and `2` non-Real Complex zeros, which are Complex conjugates of one another.

`color(white)()`
Cardano's method

Let `x = u+v`.

Then our equation becomes:

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

Multiplying out and rearranging a little:

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

Add the constraint `v=1/(3u)` to eliminate the term in `(u+v)` and get:

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

Multiply through by `27u^3` to get:

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

Using the quadratic formula, we find:

`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`

Note that these values are both real and the derivation was symmetrical in `u, v`. So we can use one of these values for `u^3` and the other for `v^3` to find the real root:

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

`color(white)(x_1) = 1/3(root(3)((27+3sqrt(69))/2)+root(3)((27-3sqrt(69))/2))`

`color(white)(x_1) ~~ 1.324717957`

The complex roots are given by multiplying `u` and `v` by suitable powers of `omega=-1/2+sqrt(3)/2i`, the primitive complex cube root of `1`...

`x_2 = 1/3(omega root(3)((27+3sqrt(69))/2)+omega^2 root(3)((27-3sqrt(69))/2))`

`x_3 = 1/3(omega^2 root(3)((27+3sqrt(69))/2)+omega root(3)((27-3sqrt(69))/2))`