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

##### 1 Answer

May 30, 2016

#### Answer:

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

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

Subtract

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

Then