How do you factor completely $64 - x^{3}$ $64-x^3$ ?

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How do you factor completely $64 - x^{3}$ $64-x^3$ ?

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Answer 1 · 2016-09-02T21:11:22

$64 - x^{3} = (4 - x) (16 + 4 x + x^{2})$ $64-x^3 = (4-x)(16+4x+x^2)$

Explanation:

The difference of cubes identity can be written:

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

Note both $64 = 4^{3}$ $64=4^3$ and $x^{3}$ $x^3$ are perfect cubes.

So we find:

$64 - x^{3} = 4^{3} - x^{3}$ $64-x^3 = 4^3-x^3$

$64 - x^{3} = (4 - x) (4^{2} + 4 x + x^{2})$ $color(white)(64-x^3) = (4-x)(4^2+4x+x^2)$

$64 - x^{3} = (4 - x) (16 + 4 x + x^{2})$ $color(white)(64-x^3) = (4-x)(16+4x+x^2)$

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How do you factor completely $64 - x^{3}$ $64-x^3$ ?

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