How do you find the values that $x$ can take by squaring? $$|x+4| \cdot |x-4| = x+4$$
How do you find the values that $x$ can take by squaring? 
How I've tried
$$|x+4|^2 \cdot |x-4|^2 = (x+4)^2$$
and
$$|x^2+16| \cdot |x^2+16| = x^2+16$$
I think I went wrong. 

Note that if $x+4 > 0$ then $|x+4| = x+4>0$ and your equation reduces to $|x-4|=1$, which implies $x=5$ or $x=3$.
Alternatively, $x+4=0$, in which case $|x+4| = x+4 =0$, so the equation holds and $x=-4$.
Finally, if $x+4 < 0$ then we get $|x-4|=-1$ which has no real solutions.
Hence, the entire set of solutions is given by $-4, 3, 5$.

If you want to find the solutions by squaring, note that $|a|^2 = a^2$, therefore squaring both sides yields
$$
(x+4)^2(x-4)^2=(x+4)^2,
$$
which if $x \ne -4$ implies  $(x-4)^2=1$, yielding $x = 3$ or $x=5$. Alternatively, if $x=-4$, you get $=0$ so the equation holds and $x=-4$ is a solution.
in summary, you get the same solution set this way.

Note, however, that in general, squaring both sides of the equation may introduce additional solutions, which are not the solutions of the original equation. Consider, for example, the equation $x=1$, which trivially has a unique solution. Squaring both sides yields $x^2=1$, for which the original $x=1$ is a solution, but the second equation has an extra solution at $x=-1$, which does not solve the original equation.
As a result, one must use such techniques very carefully, checking all the solutions you find in the process to validate that they actually solve the original equation.
