$ x-b = 4 \left | 4 | x | -b ^ 2 \right | $ has exactly three solutions. Find the sum of all values of the parameter $ b $ for which the equation $$ x-b = 4 \left | 4 | x | -b ^ 2 \right | $$ has exactly three solutions.
I tried to represent this equation in two axes and draw this graph in order to analyze the cases when there are exactly three solutions, but it comes out too cumbersome. I will be glad for any hint, thanks :)

Just reminding some basics:

*

*To plot $f(|x|)$, draw $f(x)$, ignore the left hand side part of the y-axis, reflect the graph to the right of the y-axis in the y-axis.

*To plot $|f(x)|$, draw $f(x)$, reflect any part of graph that was below the x-axis in the x-axis.

Let's draw $4|4|x| - b^2| = |16|x| - 4b^2|$:

*

*$f(x) = 16x - 4b^2$.

*$f(|x|) = 16|x| - 4b^2$

*$|f(|x|)|$ = $|16|x| - 4b^2|$

Then we have a $y = x - b$, a line parallel to lines below (or one of the lines below):

In order for $x - b = 4|4|x| - b^2|$ to have exactly three answers, $x - b$ should have three intersections with $4|4|x| - b^2|$. So $x - b$ should be one of the blue lines below and can't be any line else:

So:
$$
4b^2=-b \Rightarrow
    \begin{cases}
      b=0\\
      b=-4\\
    \end{cases} 
$$
$$
\frac{-b^2}{4}=b \Rightarrow
    \begin{cases}
      b=0\\
      b=\frac{-1}{4}\\
    \end{cases} 
$$
But if $b=0$, then $\frac{b^2}{4}=\frac{-b^2}{4}$, so the graph we drew will change and the only answer will be $x=0$.
So
$
    \begin{cases}
      b=-4
        \begin{cases}
          x=-4\\
          x=\frac{60}{17}\\
          x=\frac{68}{15}\\
        \end{cases} \\
      b=\frac{-1}{4}
        \begin{cases}
          x=\frac{-1}{34}\\
          x=0\\
          x=\frac{1}{30}\\
        \end{cases} \\
    \end{cases} 
$
