# How do you factor: y= 27x^3 - 1 ?

Jan 18, 2016

$\left(27 {x}^{3} - 1\right) = \left(3 x - 1\right) \left(9 {x}^{2} + 3 x + 1\right)$

#### Explanation:

Cube root of 27 is 3 and the cube root of -1 is also -1.
$27 {x}^{3} - 1 = {\left(3 x\right)}^{3} - {\left(1\right)}^{3}$

To factor $y = 27 {x}^{3} - 1$,
remember that $\left({a}^{3} - {b}^{3}\right) = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

substituting,

$\left(27 {x}^{3} - 1\right) = \left(3 x - 1\right) \left({\left(3 x\right)}^{2} + \left(3 x\right) \left(1\right) + {\left(1\right)}^{2}\right)$
$\left(27 {x}^{3} - 1\right) = \left(3 x - 1\right) \left(9 {x}^{2} + 3 x + 1\right)$

Jan 18, 2016

$\left(3 x - 1\right) \left(9 {x}^{2} + 3 x + 1\right)$

#### Explanation:

Notice that both the terms are cubed terms, i.e. $27 {x}^{3} = {\left(3 x\right)}^{3}$ and $1 = {1}^{3}$.

That means that this is a difference of cubes, which can be factored as follows;

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

Thus, we have $a = 3 x$ and $b = 1$:

$27 {x}^{3} - 1 = {\left(3 x\right)}^{3} - {1}^{3}$

$= \left(3 x - 1\right) \left({\left(3 x\right)}^{2} + \left(3 x\right) \left(1\right) + {1}^{2}\right)$

$= \left(3 x - 1\right) \left(9 {x}^{2} + 3 x + 1\right)$