How do we solve for all values of $x$? Suppose 
$$(x^2 - 5)^{8} (x+1)^{-16} = 1$$
How do we solve for all values of $x$?
I think we can equate the bases to $1$ in order to get $$1 \times 1 = 1$$
$$x^2-5 = 1, x^2-5=-1$$
or
$$x + 1 = 1, x + 1 = -1$$
Could you assist me with this?

We have $(x^2-5)^8(x+1)^{-16}=1$, so $\left(\frac{x^2-5}{(x+1)^2}\right)^8=1$, so $\frac{x^2-5}{(x+1)^2}$ is an $8$'th root of unity. Assuming you need to solve over the reals, we get $\frac{x^2-5}{(x+1)^2}\in\{-1,1\}$. Fixing for $-1$ or $1$, you can multiply both sides by $(x+1)^2$ to get a quadratic equation.
