Right away you know that you can't have \(x=\pm1\) as possible solutions because either of these make the first term's denominator 0, while \(x=-1\) also makes the second term's denominator 0.
With this in mind, let's fix \(x
eq\pm1\), which allows you to write
\(\dfrac8{x^2-1}+\dfrac4{x+1}=1\iff8+4(x-1)=x^2-1\)
Moving everything to one side, you have
\(x^2-4x-5=0\iff (x-5)(x+1)=0\implies x=5\text{ or }x=-1\)
We omitted \(x=-1\) from the solution set, so this is an extraneous root and we're left with a solution of \(x=5\).
