<s>[INST] 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?
 [/INST] $(x^2 - 5)^{8} (x+1)^{-16} = 1$ means
$(x^2 - 5)^{8} = (x+1)^{16}$.
Assuming you only want real and not complex answers then
$ \sqrt[8]{(x^2 - 5)^{8}} = \sqrt[8]{(x+1)^{16}}$
$|x^2 - 5| = (x+1)^2$
or either A) $x^2 - 5 = (x+1)^2$ or B) $5- x^2 = (x+1)^2$
A) $x^2 - 5 = x^2 + 2x + 1$
$2x = -6$
$x = -3$.
Verification:  $(x^2 - 5)^8(x+1)^{-16}=\frac{(9-5)^8}{(-2)^{16}}= \frac {4^8}{2^{16}}=\frac {2^{16}}{2^{16}} = 1$.
B)$5- x^2 = (x+1)^2$
$5-x^2 = x^2 + 2x + 1$
$2x^2 + 2x -4 = 0$
$x^2 + x -2 = 0$
$(x +2)(x-1) = 0$ so either C) $x+2 = 0$ and $x=-2$ or $x-1=0$ and $x = 1$.
So solutions are $x = -3; x = -2;$ or $x = 1$.
Verification:
$(x^2 - 5)^8(x+1)^{-16} = \frac {(4-5)^8}{(-2 + 1)^{16}}= \frac {(-1)^8}{(-1)^{16}} = \frac 11 = 1$. so $x = -2$ is a solution.
$(x^2 - 5)^8(x+1)^{-16} = \frac {(1-5)^8}{(1 + 1)^{16}}= \frac {(-4)^8}{(2)^{16}} = \frac {4^8}{4^8} = 1$. so $x = 1$ is a solution.
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