How do you factor $z^{3} - 4$ $z^3 - 4$ ?

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How do you factor $z^{3} - 4$ $z^3 - 4$ ?

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Answer 1 · 2016-04-26T21:35:55

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

Explanation:

This can be factored as a difference of cubes.

The difference of cubes identity can be written:

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

We can use this with $a = z$ $a=z$ and $b = \sqrt[3]{4}$ $b=root(3)(4)$ as follows:

$z^{3} - 4$ $z^3-4$

$= z^{3} - {(\sqrt[3]{4})}^{3}$ $= z^3-(root(3)(4))^3$

$= (z - \sqrt[3]{4}) (z^{2} + \sqrt[3]{4} z + \sqrt[3]{16})$ $=(z-root(3)(4))(z^2+root(3)(4)z+root(3)(16))$

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

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