Answer
$a=\{\pm i, \pm1 \}$
Work Step by Step
Putting all expression on the left side and then using factoring by grouping, the solution/s to the given equation, $
a^6-a^2=a^4-1
,$ is/are
\begin{array}{l}\require{cancel}
a^6-a^2-a^4+1=0
\\\\
(a^6-a^2)-(a^4-1)=0
\\\\
a^2(a^4-1)-(a^4-1)=0
\\\\
(a^4-1)(a^2-1)=0
\\\\
(a^2+1)(a^2-1)(a^2-1)=0
.\end{array}
Equating each factor to zero, then,
\begin{array}{l}\require{cancel}
a^2+1=0
\\\\
a^2=-1
\\\\
a=\pm\sqrt{-1}
\\\\
a=\pm i
,\\\\\text{OR}\\\\
a^2-1=0
\\\\
a^2=1
\\\\
a=\pm\sqrt{1}
\\\\
a=\pm1
.\end{array}
Hence, $
a=\{\pm i, \pm1 \}
.$
