# How do you factor completely x^3 + y^3?

Mar 30, 2018

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

$\textcolor{w h i t e}{{x}^{3} + {y}^{3}} = \left(x + y\right) \left(x - \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) y\right) \left(x - \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) y\right)$

#### Explanation:

Given:

${x}^{3} + {y}^{3}$

Note that if $y = - x$ then this is zero. So we can deduce that $\left(x + y\right)$ is a factor and separate it out:

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

We can calculate the discriminant for the remaining homogeneous quadratic in $x$ and $y$ just like we would for a quadratic in a single variable:

${x}^{2} - x y + {y}^{2}$

is in standard form:

$a {x}^{2} + b x y + c {y}^{2}$

with $a = 1$, $b = - 1$ and $c = 1$

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\textcolor{b l u e}{1}}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{1}\right) = - 3$

Since $\Delta < 0$, this quadratic has no linear factors with real coefficients.

We can factor it with complex coefficients by completing the square and using ${i}^{2} = - 1$ as follows:

${x}^{2} - x y + {y}^{2} = {\left(x - \frac{1}{2} y\right)}^{2} + \frac{3}{4} {y}^{2}$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = {\left(x - \frac{1}{2} y\right)}^{2} + {\left(\frac{\sqrt{3}}{2} y\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = {\left(x - \frac{1}{2} y\right)}^{2} - {\left(\frac{\sqrt{3}}{2} y i\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = \left(\left(x - \frac{1}{2} y\right) - \frac{\sqrt{3}}{2} i y\right) \left(\left(x - \frac{1}{2} y\right) + \frac{\sqrt{3}}{2} i y\right)$

$\textcolor{w h i t e}{{x}^{2} - x y + {y}^{2}} = \left(x - \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) y\right) \left(x - \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) y\right)$

So:

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

$\textcolor{w h i t e}{{x}^{3} + {y}^{3}} = \left(x + y\right) \left(x - \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right) y\right) \left(x - \left(\frac{1}{2} - \frac{\sqrt{3}}{2} i\right) y\right)$