## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$(2t+1)(2t-1)(4t^2-2t+1)(4t^2+2t+1)$
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}