# How do you factor a^6 + 7a^3 + 6?

Apr 15, 2016

$\left(a + 1\right) \left({a}^{2} - a + 1\right) \left({a}^{3} + 6\right)$

#### Explanation:

Call $x = {a}^{3}$. Factor the quadratic equation:
$f \left(x\right) = {x}^{2} + 7 x + 6.$
Since a - b + c = 0, use shortcut. One factor is (x + 1) and the other is $\left(x + \frac{c}{a}\right) = \left(x + 6\right)$. We get:
$f \left(x\right) = \left(x + 1\right) \left(x + 6\right)$
Replace x by a^3.
$f \left(a\right) = \left({a}^{3} + 1\right) \left({a}^{3} + 6\right)$
Factor $\left({a}^{3} + 1\right)$ and $\left({a}^{3} + 6\right)$ by applying the algebraic identity:
${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$.
Finally,
$f \left(a\right) = \left(a + 1\right) \left({a}^{2} - a + 1\right) \left({a}^{3} + 6\right)$

Apr 15, 2016

${a}^{6} + 7 {a}^{3} + 6 = \left(a + 1\right) \left({a}^{2} - a + 1\right) \left(a + \sqrt[3]{6}\right) \left(a - \sqrt[3]{6} a + \sqrt[3]{36}\right)$

#### Explanation:

First treat this as a quadratic in ${a}^{3}$, noting that $1 + 6 = 7$ and $1 \cdot 6 = 6$.

So:

${a}^{6} + 7 {a}^{3} + 6 = {\left({a}^{3}\right)}^{2} + 7 \left({a}^{3}\right) + 6 = \left({a}^{3} + 1\right) \left({a}^{3} + 6\right)$

Note that both ${a}^{3}$ and $1 = {1}^{3}$ are perfect cubes, so we can use the sum of cubes identity:

${A}^{3} + {B}^{3} = \left(A + B\right) \left({A}^{2} - A B + {B}^{2}\right)$

with $A = a$ and $B = b$ to find:

$\left({a}^{3} + 1\right) = \left({a}^{3} + {1}^{3}\right) = \left(a + 1\right) \left({a}^{2} - a + 1\right)$

The factor $\left({a}^{3} + 6\right)$ is not quite so nice, but it can still be treated as a sum of cubes, using $A = a$ and $B = \sqrt[3]{6}$ as follows:

$\left({a}^{3} + 6\right)$

$= \left({a}^{3} + {\left(\sqrt[3]{6}\right)}^{3}\right)$

$= \left(a + \sqrt[3]{6}\right) \left(a - \sqrt[3]{6} a + {\left(\sqrt[3]{6}\right)}^{2}\right)$

$= \left(a + \sqrt[3]{6}\right) \left(a - \sqrt[3]{6} a + \sqrt[3]{{6}^{2}}\right)$

$= \left(a + \sqrt[3]{6}\right) \left(a - \sqrt[3]{6} a + \sqrt[3]{36}\right)$

Putting this together:

${a}^{6} + 7 {a}^{3} + 6 = \left(a + 1\right) \left({a}^{2} - a + 1\right) \left(a + \sqrt[3]{6}\right) \left(a - \sqrt[3]{6} a + \sqrt[3]{36}\right)$

This is as far as we can go with Real numbers.