# How do I factor x3 – 27?

(The x3 is x to the third power. Not three times x)

• x^3 – 27

x^3 – 3^3

(x – 3)(x^2 + 3x + 9)

Remember these formulas,

x^3 + a^3 = (x + a)(x^2 – ax + a^2)

x^3 – a^3 = (x – a)(x^2 + ax + a^2)

2.) Not a big difference, just switch them,

27 – x^3

3^3 – x^3

(3 – x)(9 – 3x + x^2)

• Factoring X 3

• x^3 – 27 = (x – 3) (x^2 + 3x + 9)

it is an example of special products called difference of cubes..

x^3 – 27 = x^3 – 3^3

applying the solution:

x^3 – y^3 = (x – y) (x^2 + xy + y^2)

it will become…

x^3 – 3^3 = (x – 3) (x^2 + 3x + 9)

• You can factor a term like x^3 – a^3 as (x-a1)(x-a2)(x-a3), where a1, a2, a3 are the three cube roots of a^3, i.e. a, -a/2(1+i root(3)), and -a/2(1-i root(3)).

Thus, x^3-27 factors into (x-3)(x+3/2((1+i root(3))(x+3/2((1-i root(3))

As to your second question, this is just the negative of the first expression, so you multiply the factorization I gave above by -1.

Al the answers above are correct, but mine is the only complete factorization.

• The main formula is a^3 – b^ 3 = (a – b)(a^2 + ab + b^2)

We apply the same procedure here and plug in the respective values for “a” and “b”

So its: (x – 3)(x^2 + 3x + 9)

There it is…factored!

• x3 + 27 is the sum of cubes, so use the rule for factoring the sum of cubes: (x + 3)(x2 – 3x + 9)

• difference of perfect cubes

a^3 – b^3 = (a – b) (a^2 + ab + b^2)

x^3 – 3^3

(x – 3) (x^2 + 3x + 9)

• a^3 – b^3 = (a-b)(a^2+ab+b^2)

so:

x^3 – 27 = (x-3)(x^2 + 3x + 9)

• x^3 – 27

x^3 – 3^3

(x – 3)(x^2 + 3x + 9)

x^3 + a^3 = (x + a)(x^2 – ax + a^2)

x^3 – a^3 = (x – a)(x^2 + ax + a^2)

• x3-27=0

=x3-3×2+3×2-9x+9x-27

=(x-3)(x2+3x+9)

x=3