Answer
$(t^2+1)(t^4-t^2+1)$
Work Step by Step
Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $
t^6+1
,$ is
\begin{array}{l}
(t^2+1)[ (t^2)^2-(t^2)(1)+(1)^2]
\\\\=
(t^2+1)(t^4-t^2+1)
.\end{array}