# Factor completely `512x^3-1`

Asked on by jeuxeur

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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).`

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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]

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