A 6th degree?!

Alright, this is a 6th degree polynomial. One way is to substitute something for $x$ , but how do we do that?

We first expand the terms inside the outer brackets. We will get

$(x^2+x+1)(x^2+2x+1)(x^2+4x+1)=-2x^5+4x^3-2x$ $\implies \left(x+ \frac{1}{x}+1 \right) \left(x+\frac{1}{x}+2 \right) \left(x+\frac{1}{x}+4 \right)=4-2x^2-\frac{2}{x^2}$ Let $x+\frac{1}{x}=a$ . Then $(a+1)(a+2)(a+4)=-2a^2+8$ $\implies a(a+2)(a+7)=0$ $\implies x=i, -i, -1, \dfrac{-7+3\sqrt{5}}{2}, \dfrac{-7-3\sqrt{5}}{2}$

Take note that $-1$ is a double root because we will arrive with $(x+1)^2=0$ . Therefore we only count it once. Also, $i$ and $-i$ are not real numbers so they don't count.

The sum of the squares of all real values of $x$ is then $1+47=\boxed{48}$ .

It is given that:

$[(x+1)^2-x][(x+1)^2][(x+1)^2+2x] = -2x^5+4x^3-2x$

The RHS: $-2x^5+4x^3-2x = -2x(x^4-2x^2+1) = -2x(x^2-1)^2$ $\quad \quad \quad \quad = -2x(x+1)^2(x-1)^2 = -2x(x+1)^2[(x+1)^2-4x]$

Therefore,

$[(x+1)^2-x][(x+1)^2][(x+1)^2+2x] =-2x(x+1)^2[(x+1)^2-4x]$

This implies that $(x+1)^2$ is common on both sides and that $x=-1$ is a solution to the equation. It also implies that:

$[(x+1)^2-x][(x+1)^2+2x] =-2x[(x+1)^2-4x]$

$\Rightarrow (x+1)^4+x(x+1)^2-2x^2 =-2x(x+1)^2+8x$

$\Rightarrow (x+1)^4+3x(x+1)^2-10x^2 = 0$

$\Rightarrow [(x+1)^2-2x] [(x+1)^2+5x] = 0$

$\Rightarrow \begin {cases} (x+1)^2-2x = x^2+1 = 0 &\Rightarrow x = \pm i \\ (x+1)^2+5x = x^2 + 7x + 1= 0 &\Rightarrow x = \dfrac{-7\pm 3\sqrt{5}} {2} \end {cases}$

Therefore the real solutions are: $\quad -1$ , $\dfrac{-7 + 3\sqrt{5}} {2}$ and $\dfrac{-7- 3\sqrt{5}} {2}$ .

And the sum of their squares is: $\quad 1+\dfrac{47 + 21\sqrt{5}} {2} + \dfrac{47 - 21\sqrt{5}} {2} = \boxed{48}$