Work It Harder, Make It Better

First we want to find the roots of the equation, a, b, c.

To do this we will expand the equation and then factorise.

The equation expands to $2{ x }^{ 3 }+6{ x }^{ 2 }-12x-16$

Factorising this we get the equation $2(x-2)(x+1)(x+4)$

The roots of this equation are $x=2$ , $x=-1$ , $x=-4$

Therefore $a=2$ , $b=-1$ , $c=-4$

Substituting into the equation

$\begin{aligned} \dfrac{\left (\dfrac{a+b}{a^a+b^b} \right )^{(a+\sqrt{bc}+c)}}{\left (\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c} \right )^{(a-b+c)}} \end{aligned}$

$\begin{aligned} \dfrac{\left (\dfrac{2-1}{2^2+-1^{-1}} \right )^{(2+\sqrt{-1(-4)}+c)}}{\left (\dfrac{1}{2-1}+\dfrac{1}{2-4}+\dfrac{1}{-1-4} \right )^{(2+1-4)}} \end{aligned}$

Which then simplifies to $\frac { 1 }{ \frac { 10 }{ 3 } }$

$\frac { 10 }{ 3 } =\frac { m }{ n }$

Substituting into the equation

$\begin{aligned} \dfrac{\sqrt{m^m+n^m-m}}{3m^m-n^2+20}.\end{aligned}$

$\begin{aligned} \dfrac{\sqrt{10^{10}+3^{10}-10}}{3(10^{10})-3^{2}+20}.\end{aligned}$

Which simplifies to $\sqrt { 1024 }$

$\sqrt { 2^{10} }$

${ 2 }^{ 5 }$

= $\boxed{32}$