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

May 30, 2016

${x}^{3} + x - 3 = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$

where:

${x}_{1} = \frac{1}{3} \left(\sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}}\right)$

etc.

#### Explanation:

There we find zeros:

${x}_{1} = \frac{1}{3} \left(\sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}}\right)$

${x}_{2} = \frac{1}{3} \left(\omega \sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + {\omega}^{2} \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}}\right)$

${x}_{3} = \frac{1}{3} \left({\omega}^{2} \sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + \omega \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ 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 ${x}^{3} + x - 3 = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \left(x - {x}_{3}\right)$