Hi Lizzie;

You are obviously working on the premise that only x^{2} can be factored. This is incorrect. A factor can be...

(x^{2} )(x )

or x^{3}, x^{4}, etc.

Henceforth

27x^{3}-1

is factorable.

To factor, we must do the following...

We know that only

1x1=1

Henceforth...

(x^{2} 1)(x 1)

We also know that one of the 1s must be negative to yield -1. Therefore, one of the parenthetical equations must engage in subtraction, while the other addition.

To yield 27...

9 x 3

is the only possibility, and both numbers must either be negative or positive.

We also know that the FOIL equation must yield 0x. The above will produce 1x^{2} and 1x.

So we must start again as follows...

(x 1)(x 1)

is the only equation which will work. However, we cannot forget about the third x. Therefore...

x(x + 1)(x - 1)

We also cannot forget about the 27. Furthermore, both parenthetical x-s must be multiplied by the same number to yield 0x.

3x3x3 is the only mathematical formula which will yield 27.

3x(3x+1)(3x-1)

FOIL

FIRST...3x(9x^{3})

OUTER...3x(-3x)

INNER...3x(3x)

LAST...-1

27x^{3}-9x^{2}+9x^{2}-1=27x^{3}-1

Please let me know if you need anything else.

I LOVE MATH!!!!