How do you factor $x^{3} - 16$ $x^3 - 16$ ?

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How do you factor $x^{3} - 16$ $x^3 - 16$ ?

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Answer 1 · 2016-04-07T22:21:01

$x^{3} - 16$ $x^3-16$

$= (x - 2 \sqrt[3]{2}) (x^{2} + 2 \sqrt[3]{2} x + 4 \sqrt[3]{4})$ $=(x-2 root(3)(2))(x^2+2 root(3)(2) x + 4 root(3)(4))$

$= (x - 2 \sqrt[3]{2}) (x - 2 \sqrt[3]{2} ω) (x - 2 \sqrt[3]{2} ω^{2})$ $=(x-2 root(3)(2))(x-2 root(3)(2) omega)(x-2 root(3)(2) omega^2)$

Explanation:

Use the difference of cubes identity:

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

with $a = x$ $a = x$ and $b = 2 \sqrt[3]{2}$ $b = 2root(3)(2)$

So:

$x^{3} - 16 = x^{3} - {(2 \sqrt[3]{2})}^{3}$ $x^3-16 =x^3-(2 root(3)(2))^3$

$= (x - 2 \sqrt[3]{2}) (x^{2} + 2 \sqrt[3]{2} x + 4 \sqrt[3]{4})$ $=(x-2 root(3)(2))(x^2+2 root(3)(2) x + 4 root(3)(4))$

This is as far as we can go with Real coefficients, but if you allow Complex coefficients then this can be further factored as:

$= (x - 2 \sqrt[3]{2}) (x - 2 \sqrt[3]{2} ω) (x - 2 \sqrt[3]{2} ω^{2})$ $=(x-2 root(3)(2))(x-2 root(3)(2) omega)(x-2 root(3)(2) omega^2)$

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 $x^{3} - 16$ $x^3 - 16$ ?

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