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?

$(x^2-5)^8=(x+1)^{16}$
So
$[(x^2-5)^4-(x+1)^8] [(x^2-5)^4+(x+1)^8]=0$
If and only if
$[(x^2-5)^4-(x+1)^8]=0$
If and only if
$[(x^2-5)^2-(x+1)^4]=0$
If and only if 
$[(x^2-5)-(x+1)^2] [(x^2-5)+(x+1)^2]=0$
So
$[-2x-6][2x^2+2x-4]=0$
