# How do you factor 27y^6-8?

Rewrite this as

$27 \cdot {y}^{6} - 8 = {\left(3 {y}^{2}\right)}^{3} - {2}^{3}$

Using the cubes difference formula

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

where $a = 3 {y}^{2}$ and $b = 2$

we get

$27 {y}^{6} - 8 = \left(3 {y}^{2} - 2\right) \left(9 {y}^{4} + 6 {y}^{2} + 4\right)$

But this can be factored further as follows

$27 {y}^{6} - 8 = \frac{1}{3} \left(\sqrt{6} - 3 y\right) \left(3 y + \sqrt{6}\right) \left(- 3 {y}^{2} + \sqrt{6} y - 2\right) \left(3 {y}^{2} + \sqrt{6} y + 2\right)$