# How do you factor 13(x^6+ 1)^4(18x^5)(9x + 2)^3 + 9(9x + 2)^2(9)(x^6 + 1)^5?

Apr 15, 2016

$13 {\left({x}^{6} + 1\right)}^{4} \left(18 {x}^{5}\right) {\left(9 x + 2\right)}^{3} + 9 {\left(9 x + 2\right)}^{2} \left(9\right) {\left({x}^{6} + 1\right)}^{5}$

$= 9 {\left({x}^{2} + 1\right)}^{4} {\left({x}^{2} - \sqrt{3} x + 1\right)}^{4} {\left({x}^{2} + \sqrt{3} x + 1\right)}^{4} {\left(9 x + 2\right)}^{2} \left(243 {x}^{6} + 52 {x}^{5} + 9\right)$

#### Explanation:

Given:

$13 {\left({x}^{6} + 1\right)}^{4} \left(18 {x}^{5}\right) {\left(9 x + 2\right)}^{3} + 9 {\left(9 x + 2\right)}^{2} \left(9\right) {\left({x}^{6} + 1\right)}^{5}$

First note that both of the terms are divisible by:

$9 {\left({x}^{6} + 1\right)}^{4} {\left(9 x + 2\right)}^{2}$

So separate that out as a factor first to get:

$13 {\left({x}^{6} + 1\right)}^{4} \left(18 {x}^{5}\right) {\left(9 x + 2\right)}^{3} + 9 {\left(9 x + 2\right)}^{2} \left(9\right) {\left({x}^{6} + 1\right)}^{5}$

$= 9 {\left({x}^{6} + 1\right)}^{4} {\left(9 x + 2\right)}^{2} \left(13 \left(2 {x}^{5}\right) \left(9 x + 2\right) + \left(9\right) \left({x}^{6} + 1\right)\right)$

$= 9 {\left({x}^{6} + 1\right)}^{4} {\left(9 x + 2\right)}^{2} \left(243 {x}^{6} + 52 {x}^{5} + 9\right)$

We can treat ${x}^{6} + 1$ as a sum of cubes to factorise it some way:

The sum of cubes identity can be written:

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

Use this with $a = {x}^{2}$ and $b = 1$ to find:

${x}^{6} + 1 = {\left({x}^{2}\right)}^{3} + {1}^{3} = \left({x}^{2} + 1\right) \left({x}^{4} - {x}^{2} + 1\right)$

Next note that:

$\left({a}^{2} - k a b + {b}^{2}\right) \left({a}^{2} + k a b + {b}^{2}\right) = \left({a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4}\right)$

So, putting $a = x$, $b = 1$ and $k = \sqrt{3}$ we find:

${x}^{4} - {x}^{2} + 1 = \left({x}^{2} - \sqrt{3} x + 1\right) \left({x}^{2} + \sqrt{3} x + 1\right)$

So:

${x}^{6} + 1 = \left({x}^{2} + 1\right) \left({x}^{2} - \sqrt{3} x + 1\right) \left({x}^{2} + \sqrt{3} x + 1\right)$

Putting it all together:

$13 {\left({x}^{6} + 1\right)}^{4} \left(18 {x}^{5}\right) {\left(9 x + 2\right)}^{3} + 9 {\left(9 x + 2\right)}^{2} \left(9\right) {\left({x}^{6} + 1\right)}^{5}$

$= 9 {\left({x}^{2} + 1\right)}^{4} {\left({x}^{2} - \sqrt{3} x + 1\right)}^{4} {\left({x}^{2} + \sqrt{3} x + 1\right)}^{4} {\left(9 x + 2\right)}^{2} \left(243 {x}^{6} + 52 {x}^{5} + 9\right)$

The remaining quadratic factors have no Real zeros and therefore no linear factors with Real coefficients.

The remaining sextic factor has $6$ Complex zeros in $3$ conjugate pairs. That means that in theory it could be factored into $3$ quadratic factors with Real coefficients, but as far as I can tell the coefficients of those factors are not expressible in terms of $n$th roots and other ordinary arithmetic operations. That is, they are not determinable by practical algebraic means. It is possible to find numeric approximations using Newton Raphson, Durand-Kerner or other similar numeric methods.