Question

How do you solve `x^ { 3} - 6x = 12`?

Answer 1

Use Cardano's method to find real root:

`x = root(3)(6+2sqrt(7))+root(3)(6-2sqrt(7))`

and related complex roots.

Explanation:

Given:

`x^3-6x=12`

Subtract `12` from both sides to get the standard form:

`x^3-6x-12 = 0`

This cubic equation has one irrational real root and two complex roots, which we can find using Cardano's method.

Letting `x=u+v`, our equation becomes:

`u^3+v^3+3(uv-2)(u+v)-12 = 0`

Add the constraint `(uv-2) = 0` to eliminate the term in `(u+v)`. Then our equation becomes:

`u^3+8/u^3-12 = 0`

Multiply through by `u^3` and rearrange slightly to find:

`0 = (u^3)^2-12(u^3)+8`

`color(white)(0) = (u^3)^2-2(u^3)(6)+36-28`

`color(white)(0) = (u^3-6)^2-(2sqrt(7))^2`

`color(white)(0) = ((u^3-6)-2sqrt(7))((u^3-6)+2sqrt(7))`

`color(white)(0) = (u^3-6-2sqrt(7))(u^3-6+2sqrt(7))`

That is:

`u^3 = 6+-2sqrt(7)`

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

`x_1 = root(3)(6+2sqrt(7))+root(3)(6-2sqrt(7))`

and related complex roots:

`x_2 = omega root(3)(6+2sqrt(7))+omega^2 root(3)(6-2sqrt(7))`

`x_3 = omega^2 root(3)(6+2sqrt(7))+omega root(3)(6-2sqrt(7))`

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