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

##### 2 Answers

See below.

#### Explanation:

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

You can rewrite the equation as

Then graph the following:

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

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

#x^3-x-1 = 0#

**Discriminant**

The discriminant

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

In our example,

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

Since

**Cardano's method**

Let

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

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

Multiply through by

#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

#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

#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))#