How do you factor $(r+6)^3-216$ ?

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Answer 1 · 2016-05-02T07:53:41

$(r+6)^3-216=r(r^2+18r+108)$

The difference of cubes identity can be written:

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

We can use this with $a=(r+6)$ and $b=6$ as follows:

$(r+6)^3-216$

$=(r+6)^3-6^3$

$=((r+6)-6)((r+6)^2+(r+6)(6)+6^2)$

$=r((r^2+12r+36)+(6r+36)+36)$

$=r(r^2+18r+108)$

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Footnote

The remaining quadratic factor cannot be factorised further using Real coefficients.

It can be factorised by completing the square using Complex coefficients:

$r^2+18r+108$

$=(r+9)^2-81+108$

$=(r+9)^2+27$

$=(r+9)^2-(3sqrt(3)i)^2$

$=((r+9)-3sqrt(3)i)((r+9)+3sqrt(3)i)$

$=(r+9-3sqrt(3)i)(r+9+3sqrt(3)i)$

How do you factor (r+6)^3-216(r+6)^3-216?