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

$(x^4+16)(x^2+4)(x+2)(x-2)$
Using $x^2-y^2=(x+y)(x-y)$ or the factoring of the difference of 2 squares, then the factored form of the given expression, $x^8-2^8 ,$ is \begin{array}{l} (x^4+2^4)(x^4-2^4) \\\\= (x^4+2^4)(x^2+2^2)(x^2-2^2) \\\\= (x^4+2^4)(x^2+2^2)(x+2)(x-2) \\\\= (x^4+16)(x^2+4)(x+2)(x-2) \end{array}