Find all $a$ for which the equation $8x^6+(a-|x|)^3+2x^2-|x|+a=0$ has more than $3$ different roots.

Question

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.

Answer 1

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.