# How do you factor 8a^3+27(x+y)^3?

Apr 11, 2015

$\left(2 a + 3 x + 3 y\right) \left(4 {a}^{2} - 6 a x - 6 a y + 9 {x}^{2} + 18 x y + 9 {y}^{2}\right)$

If you write:

$8 {a}^{3} + 27 {\left(x + y\right)}^{3} = {\left(2 a\right)}^{3} + {\left[3 \left(x + y\right)\right]}^{3} = \left(1\right)$

This is a sum of two cubic, and you can use this rule:

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

So:

$\left(1\right) = \left[2 a + 3 \left(x + y\right)\right] \left[{\left(2 a\right)}^{2} - \left(2 a\right) \cdot 3 \left(x + y\right) + {3}^{2} {\left(x + y\right)}^{2}\right] =$

$= \left(2 a + 3 x + 3 y\right) \left(4 {a}^{2} - 6 a x - 6 a y + 9 {x}^{2} + 18 x y + 9 {y}^{2}\right)$.

Apr 11, 2015

Temporarily simplify be replacing
$8 {a}^{3} = {\left(2 a\right)}^{3}$ with ${p}^{3}$
and
$27 {\left(x + y\right)}^{3} = {\left(3 \left(x + y\right)\right)}^{3}$ with ${q}^{3}$

So $8 {a}^{3} + 27 {\left(x + y\right)}^{3}$
becomes ${p}^{3} + {q}^{3}$
which has factors
$\left(p + q\right) \left({p}^{2} - p q + {q}^{2}\right)$

Restoring the original values, this becomes
$\left(2 a + 3 \left(x + y\right)\right) \cdot \left(4 {a}^{2} - 6 a \left(x + y\right) + 9 {\left(x + y\right)}^{2}\right)$

It might be tempting to attempt to further factor the right-most term but a quick check using the formula for roots
$\frac{- b \pm s q r \left({b}^{2} - 4 a c\right)}{2} a$ reveals that no Real roots are possible
[be careful not to confuse the $a$ in this general formula with the $a$ in the example expression]