# How do you factor completely 27+8t^3?

$27 + 8 {t}^{3} = \left(3 + 2 t\right) \left(9 - 6 t + 4 {t}^{2}\right)$
$27 , 8 \mathmr{and} {t}^{3}$ are all cubes. Once you recognise that fact, you can factorise this expression by the general rule for "Sum of cubes".
${x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right)$
$27 + 8 {t}^{3} = \left(3 + 2 t\right) \left(9 - 6 t + 4 {t}^{2}\right)$