I am being asked to "Factor Completely" this question: 27x

^{3}-1Is this possible?

I am being asked to "Factor Completely" this question: 27x^{3}-1

Is this possible?

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Oconomowoc, WI

Yes it is possible.

Actually both of the terms in the equation are cubed.

27x^{3} = (3x)^{3 } and (-1) = (-1)^{3}= (-1)(-1)(-1) = 1(-1) = -1 or (1)^{3} = 1

realizing this we can use formula for (a^{3} - b^{3}) = (a-b)(a^{2} + ab + b^{2})

in your example

- a = 3x
- b = 1

substituting these values in

(3x)^{3} - (1)^{3} = (3x - 1)((3x)^{2} + (3x)(1) + 1^{2}) = (3x -1) (9x^{2} + 3x + 1)

9x^{2} + 3x + 1 can't be factored any more so your answer is (3x -1)(9x^{2} + 3x + 1)

to validate this answer do this by multiply out the answer

9x^{2} + 3x + 1

* 3x - 1

---------------

27x^{3} + 9x^{2} + 3x

+ (-9x^{2} - 3x - 1)

--------------------------

27x^{3} + 0x^{2} + 0x -1 ⇒ 27x^{3} -1

Hope this helps

Middletown, CT

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!!!!

Jane S.

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