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

May 2, 2016

${\left(r + 6\right)}^{3} - 216 = r \left({r}^{2} + 18 r + 108\right)$

#### Explanation:

The difference of cubes identity can be written:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

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

${\left(r + 6\right)}^{3} - 216$

$= {\left(r + 6\right)}^{3} - {6}^{3}$

$= \left(\left(r + 6\right) - 6\right) \left({\left(r + 6\right)}^{2} + \left(r + 6\right) \left(6\right) + {6}^{2}\right)$

$= r \left(\left({r}^{2} + 12 r + 36\right) + \left(6 r + 36\right) + 36\right)$

$= r \left({r}^{2} + 18 r + 108\right)$

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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:

${r}^{2} + 18 r + 108$

$= {\left(r + 9\right)}^{2} - 81 + 108$

$= {\left(r + 9\right)}^{2} + 27$

$= {\left(r + 9\right)}^{2} - {\left(3 \sqrt{3} i\right)}^{2}$

$= \left(\left(r + 9\right) - 3 \sqrt{3} i\right) \left(\left(r + 9\right) + 3 \sqrt{3} i\right)$

$= \left(r + 9 - 3 \sqrt{3} i\right) \left(r + 9 + 3 \sqrt{3} i\right)$