Answer
$(2t+1)(2t-1)(4t^2-2t+1)(4t^2+2t+1)$
Work Step by Step
Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, then the factored form of the given expression, $
64t^6-1
$, is
\begin{array}{l}
(8t^3+1)(8t^3-1)
.\end{array}
Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of $2$ cubes, then the factored form of the expression, $
(8t^3+1)(8t^3-1)
$ is
\begin{array}{l}
(2t+1)[(2t)^2-(2t)(1)+(1)^2](2t-1)[(2t)^2-(2t)(-1)+(-1)^2]
\\\\=
(2t+1)(4t^2-2t+1)(2t-1)(4t^2+2t+1)
\\\\=
(2t+1)(2t-1)(4t^2-2t+1)(4t^2+2t+1)
.\end{array}
