# Factor completely `512x^3-1`

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### 2 Answers

You should use the formula that factors a difference of two cubes such that:

`x^3 - y^3 = (x^2 + xy + y^2)(x - y)`

Considering `x = root(3)(512x^3) = 8x` and `y = root(3)1 = 1` yields:

`512x^3 - 1 = ((8x^2) + 8x*1 + 1^2)(8x - 1)`

`512x^3 - 1 = (64x^2 + 8x + 1)(8x - 1)`

Since the expression `64x^2 + 8x + 1 > 0` , hence, there are no real x values for the expression to be converted in factored form.

**Hence, supposing that you need to use real coefficients in the factored form, you may factor the given expression such that **`512x^3 - 1 = (64x^2 + 8x + 1)(8x - 1).`

We notice that 512 = 8^3 and we'll re-write the given expression;

512x^3-1 = (8*x)^3 - 1^3

We'll write the formula of the difference of cubes:

a^3 - b^3 = (a-b)(a^2 + ab + b^2)

We'll put a = 8x and b = 1.

(8*x)^3 - 1^3 = (8x - 1)(64x^2 + 8x + 1)

We'll determine the roots ofÂ the quadratic: 64x^2 + 8x + 1

64x^2 + 8x + 1 = 0

We'll apply quadratic formula:

x1 = [-8+sqrt(64 - 256)]/2*64

x1 = (-8+6isqrt3)/2*64

x1 = (-4+3isqrt3)/64

x2 = (-4-3isqrt3)/64

x1 and x2 are imaginary roots.

64x^2 + 8x + 1 = [x - (-4+3isqrt3)/64][x - (-4-3isqrt3)/64]

The given expression, factored completely is:

**(8x)^3 - 1^3 = (8x - 1)[x - (-4+3isqrt3)/64][x - (-4-3isqrt3)/64]**