How do you factor: y= x^3 - 27 ?

May 5, 2016

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

Explanation:

Both ${x}^{3}$ and $27 = {3}^{3}$ are perfect cubes. So it is natural to use the difference of cubes identity:

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

with $a = x$ and $b = 3$ as follows:

${x}^{3} - 27$

$= {x}^{3} - {3}^{3}$

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

$= \left(x - 3\right) \left({x}^{2} + 3 x + 9\right)$