 Katherine Walls

2021-12-22

How do you factor ${x}^{3}-27$ ? twineg4

Both ${x}^{3}$ and $27={3}^{3}$ are perfect cubes. So we can use the difference of cubes identity:
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$
with $a=x$ $b=3$ as follows:
$x-27={x}^{3}-{3}^{3}$
$=\left(x-3\right)\left({x}^{2}+x\left(3\right)+{3}^{2}\right)$
$=\left(x-3\right)\left({x}^{2}+3x+9\right)$
This is as far as you can go with Real coefficients. If you allow Complex coefficients then you can factor this a little further:
$=\left(x-3\right)\left(x-3\omega \right)\left(x-3{\omega }^{2}\right)$
where $\omega =-\frac{12}{+}\frac{\sqrt{3}}{2}i$ is the primitive Complex cube root of 1 alexandrebaud43

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