# How do you factor (2c+3)^3 - (c-4)^3?

Apr 11, 2015

Temporarily simplify by letting
$p = \left(2 c + 3\right)$
and
$q = \left(c - 4\right)$

${\left(2 c + 3\right)}^{3} - {\left(c - 4\right)}^{3}$
becomes
${p}^{3} - {q}^{3}$
which factors as
$\left(p - q\right) \left({p}^{2} + p q + {q}^{2}\right)$

Re-inserting the original values for $p$ and $q$
(2c+3-(c-4))*( (2c+3)^2 +(2c+3)(c-4) +(c-4)^2))

$= \left(c + 7\right) \cdot \left(\left(4 {c}^{2} + 12 c + 9\right) + \left(2 {c}^{2} - 5 c - 12\right) + \left({c}^{2} - 8 c + 16\right)\right)$

$= \left(c + 7\right) \left(7 {c}^{2} - c + 13\right)$
...assuming I've been able to keep all my terms straight