the x cubed part is throwing me off
Well if $\displaystyle \ln\left(x^2-1\right)\left(x^3-16x\right) = 0$ then either $\displaystyle \ln\left(x^2-1\right) = 0$ or $\displaystyle x^3-16x = 0$.
You factored both $\displaystyle x^2-1$ and $\displaystyle x^3-16x$ right! So what are the solutions for $\displaystyle x^3-16x = 0$?
For the other one, remember it's under the $\displaystyle \ln$ operator, so factoring won't do anything for now.
You need to solve $\displaystyle \ln(x^2-1) = 0,$ so remember that $\displaystyle \ln(1) = 0$ and so set $\displaystyle x^2-1 = 1$.
Sorry, I'm not following here. Could you explain what you're solving? If it's solving $\displaystyle x^2-1 = 1$, then
$\displaystyle (x-1)(x+1) = 1$ does not imply that either $\displaystyle (x-1) = 1$ or $\displaystyle (x+1) = 1$. Also, see the note.
You were solving x(x-4)(x+4) = 0 and you got x = 4 or x = -4 as solutions. What's the other solution?