Question

How do you factor `x^3 + x = 3`?

Answer 1

`x^3+x-3 = (x-x_1)(x-x_2)(x-x_3)`

where:

`x_1 = 1/3(root(3)((81+3sqrt(741))/2)+root(3)((81-3sqrt(741))/2))`

etc.

Explanation:

See:
How do you find all the real and complex roots of `f(x) = x^3+x-3`?

There we find zeros:

`x_1 = 1/3(root(3)((81+3sqrt(741))/2)+root(3)((81-3sqrt(741))/2))`

`x_2 = 1/3(omega root(3)((81+3sqrt(741))/2)+omega^2 root(3)((81-3sqrt(741))/2))`

`x_3 = 1/3(omega^2 root(3)((81+3sqrt(741))/2)+omega root(3)((81-3sqrt(741))/2))`

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

Subtract `3` from both sides of the given equation to get:

`x^3+x-3 = 0`.

Then `x^3+x-3 = (x-x_1)(x-x_2)(x-x_3)`