Chapter 5 - Polynomials and Factoring - 5.6 Factoring: A General Strategy - 5.6 Exercise Set - Page 344: 79

$(1+4x^{6}y^{6})(1+2x^{3}y^{3})(1-2x^{3}y^{3})$

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, $ 1-16x^{12}y^{12} $, is \begin{array}{l} (1+4x^{6}y^{6})(1-4x^{6}y^{6}) \\\\= (1+4x^{6}y^{6})(1+2x^{3}y^{3})(1-2x^{3}y^{3}) .\end{array}