(The x3 is x to the third power. Not three times x)
What about 27x^3?
12 Answers

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 (xa1)(xa2)(xa3), where a1, a2, a3 are the three cube roots of a^3, i.e. a, a/2(1+i root(3)), and a/2(1i root(3)).
Thus, x^327 factors into (x3)(x+3/2((1+i root(3))(x+3/2((1i 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 = (ab)(a^2+ab+b^2)
so:
x^3 – 27 = (x3)(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)

x327=0
=x33×2+3×29x+9x27
=(x3)(x2+3x+9)
x=3