How do you factor x3−27 ?

Both ${\displaystyle x^3}$ and ${\displaystyle 27=3^3}$ are perfect cubes. So we can use the difference of cubes identity:
${\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)}$
with ${\displaystyle a=x}$ ${\displaystyle b=3}$ as follows:
${\displaystyle x-27=x^3-3^3}$
${\displaystyle =(x-3)(x^2+x\left(3\right)+3^2)}$
${\displaystyle =(x-3)(x^2+3x+9)}$
This is as far as you can go with Real coefficients. If you allow Complex coefficients then you can factor this a little further:
${\displaystyle =(x-3)(x-3\omega )(x-3{\omega }^2)}$
where ${\displaystyle \omega =-\frac{12}{+}\frac{\sqrt{3}}{2}i}$ is the primitive Complex cube root of 1

Rewrite 27 as ${\displaystyle 3^3}$
${\displaystyle x^3-3^3}$
Since both terms are perfect cubes, factor using the difference of cubes formula, ${\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)}$ where ${\displaystyle a=x}$ and ${\displaystyle b=3}$
${\displaystyle (x-3)(x^2+x\cdot 3+3^2)}$
Simplify.
Move 3 to the left of x.
${\displaystyle (x-3)(x^2+3x+3^2)}$
Raise 3 to the power of 2.
${\displaystyle (x-3)(x^2+3x+9)}$

Very detailed answer. Thanks.