Answer
$x=\{\pm i, \pm1 \}$
Work Step by Step
Using factoring by grouping, the given equation, $
x^6+1=x^4+x^2
$, is equivalent to
\begin{array}{l}\require{cancel}
x^6+1-x^4-x^2=0
\\\\
(x^6-x^4)-(x^2-1)=0
\\\\
x^4(x^2-1)-(x^2-1)=0
\\\\
(x^2-1)(x^4-1)=0
\\\\
(x^2-1)(x^2-1)(x^2+1)=0
.\end{array}
Equating each factor to $0$, then,
\begin{array}{l}\require{cancel}
x^2-1=0
\\\\
x^2=1
\\\\
x=\pm\sqrt{1}
\\\\
x=\pm1
,\\\\\text{OR}\\\\
x^2+1=0
\\\\
x^2=-1
\\\\
x=\pm\sqrt{-1}
\\\\
x=\pm i
.\end{array}
Hence, $
x=\{\pm i, \pm1 \}
.$
