Katherine Walls

2021-12-22

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

twineg4

Beginner2021-12-23Added 33 answers

Both $x}^{3$ and $27={3}^{3}$ are perfect cubes. So we can use the difference of cubes identity:

${a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2})$

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

$x-27={x}^{3}-{3}^{3}$

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

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

This is as far as you can go with Real coefficients. If you allow Complex coefficients then you can factor this a little further:

$=(x-3)(x-3\omega )(x-3{\omega}^{2})$

where$\omega =-\frac{12}{+}\frac{\sqrt{3}}{2}i$ is the primitive Complex cube root of 1

with

This is as far as you can go with Real coefficients. If you allow Complex coefficients then you can factor this a little further:

where

alexandrebaud43

Beginner2021-12-24Added 36 answers

Rewrite 27 as $3}^{3$

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

Since both terms are perfect cubes, factor using the difference of cubes formula,${a}^{3}-{b}^{3}=(a-b)({a}^{2}+ab+{b}^{2})$ where $a=x$ and $b=3$

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

Simplify.

Move 3 to the left of x.

$(x-3)({x}^{2}+3x+{3}^{2})$

Raise 3 to the power of 2.

$(x-3)({x}^{2}+3x+9)$

Since both terms are perfect cubes, factor using the difference of cubes formula,

Simplify.

Move 3 to the left of x.

Raise 3 to the power of 2.

karton

Skilled2021-12-30Added 439 answers

Very detailed answer. Thanks.