How do you factor $27 x^{3} - 512$ $27x^3-512$ ?

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How do you factor $27 x^{3} - 512$ $27x^3-512$ ?

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Answer 1 · 2015-12-17T20:13:32

Use the difference of cubes identity to find:

$27 x^{3} - 512 = (3 x - 8) (9 x^{2} + 24 x + 64)$ $27x^3-512 = (3x-8)(9x^2+24x+64)$

Explanation:

The difference of cubes identity may be written:

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3=(a-b)(a^2+ab+b^2)$

Notice that $27 x^{3} = {(3 x)}^{3}$ $27x^3 = (3x)^3$ and $512 = 8^{3}$ $512 = 8^3$ are both perfect cubes. So let $a = 3 x$ $a=3x$ and $b = 8$ $b=8$ to find:

$27 x^{3} - 512$ $27x^3-512$

$= {(3 x)}^{3} - 8^{3}$ $=(3x)^3-8^3$

$= (3 x - 8) ({(3 x)}^{2} + (3 x) (8) + 8^{2})$ $=(3x-8)((3x)^2+(3x)(8)+8^2)$

$= (3 x - 8) (9 x^{2} + 24 x + 64)$ $=(3x-8)(9x^2+24x+64)$

If you allow Complex coefficients then this factors a little further:

$= (3 x - 8) (3 x - 8 ω) (3 x - 8 ω^{2})$ $=(3x-8)(3x-8omega)(3x-8omega^2)$

where $ω = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $omega = -1/2+sqrt(3)/2 i$ is the primitive Complex cube root of $1$ $1$ .

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How do you factor $27 x^{3} - 512$ $27x^3-512$ ?

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