# How do you factor 6x^3+1?

May 14, 2015

There is a useful equality:
${a}^{n} - {b}^{n} =$
$= \left(a - b\right) \cdot \left({a}^{n - 1} {b}^{0} + {a}^{n - 2} {b}^{1} + {a}^{n - 3} {b}^{2} + \ldots + {a}^{2} {b}^{n - 3} + {a}^{1} {b}^{n - 2} + {a}^{0} {b}^{n - 1}\right)$

It can easily be proven by induction.
In case of $n = 3$ it looks simple:
${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

Using this equality for $n = 3$, $a = {6}^{\frac{1}{3}} x$ and $b = - 1$, we get
$6 {x}^{3} + 1 = \left({6}^{\frac{1}{3}} x + 1\right) \left({6}^{\frac{2}{3}} {x}^{2} - {6}^{\frac{1}{3}} x + 1\right)$