Chapter 8 - Polynomials and Factoring - 8-4 Multiplying Special Cases - Practice and Problem-Solving Exercises - Page 496: 43

$p^{8}-18p^{4}q^{2}+81q^{4}$

$(p^{4}-9q^{2})^{2}$ Using the rule of $(a-b)^{2}$ = $a^{2}$-2ab+$b^{2}$ In this case, the a= $p^{4}$ and b= $-9q^{2}$ as substituting these gives us the original polynomial $a^{2}$-2ab+$b^{2}$ $(p^{4})^{2}-2(p^{4})(-9q^{2})+(-9q^{2})^{2}$ $p^{8}-18p^{4}q^{2}+81q^{4}$