# How do you factor 8a^3 + 1?

$\left(2 a + 1\right) \left(4 {a}^{2} - 2 a + 1\right)$
${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$
$8 {a}^{3} + 1 = {\left(2 a\right)}^{3} + {\left(1\right)}^{3} = \left(2 a + 1\right) \left(4 {a}^{2} - 2 a + 1\right)$