Find all $a$ for which the equation $8x^6+(a-|x|)^3+2x^2-|x|+a=0$ has more than $3$ different roots. Find all $a$ for which the equation $8x^6+(a-|x|)^3+2x^2-|x|+a=0$ has more than $3$ different roots.
I found couple of important things:
First a little rearrangement: $8|x|^6+(a-|x|)^3+2|x|^2-|x|+a=0$ and we notice that the left side is even function. The graph is symmetric, it is useful in determining the number of roots.
Second rearrangement: $(2|x|^2)^3+(a-|x|)^3+2|x|^2+(a-|x|)=0$ now I tried sum of cubes. But
I can't see how it helps me.
Hope you will help me to see what I am missing.
 A: As you have noticed, the expression is symmetric in $x$. To get rid of the absolute value signs, we consider non-negative roots of the polynomial
$$
p(x)=8x^6+(a-x)^3+2x^2-x+a.
$$
I tried to find a second degree polynomial that divides $p$, and finally found that  $x^2-\frac{1}{2}x+\frac{a}{2}$ does the job (indeed, this is the weak point of this solution see below, however). A polynomial division gives
$$
p(x)=\Bigl(x^2-\frac{1}{2}x+\frac{a}{2}\Bigr)\Bigl(8x^4+4x^3+2(1-2a)x^2-4ax+2(1+a^2)\Bigr)
$$
Completing the square in the second parenthesis (in $a$, not in $x$!),
$$
\Bigl(8x^4+4x^3+2(1-2a)x^2-4ax+2(1+a^2)\Bigr)=2\Bigl(\bigl(a-(x+x^2)\bigr)^2+1+3x^4\Bigr)\geq 2
$$
for all $x$ and $a$. It remains to check the second degree polynomial $x^2-\frac{1}{2}x+\frac{a}{2}$ for nonnegative zeros. 
I leave the last step to you, but unless I do it wrongly, the final answer is that your equation has more than three distinct roots precisely when $0<a<1/8$.
Update
A note about the weak point above (or, if you want, a different way to solve this problem):
You write that you have found that
$$
(2|x|^2)^3+(a-|x|)^3+2|x|^2+(a-|x|)=0.
$$
This is on the form
$$
A^3+B^3+A+B=0
$$
This is certainly fulfilled if $A=-B$, that is
$$
2|x|^2=|x|-a.
$$
This fills the detail about the weak point in the solution above.
