Chapter 5 - Polynomials and Factoring - 5.4 Factoring Perfect-Square Trinomials and Differences of Squares - 5.4 Exercise Set - Page 333: 113

$(3p+5)(3p-5)^2$

Using factoring by grouping, the factored form of the given expression, $ 27p^3-45p^2-75p+125 ,$ is \begin{array}{l}\require{cancel} (27p^3-45p^2)-(75p-125) \\\\= 9p^2(3p-5)-25(3p-5) \\\\= (3p-5)(9p^2-25) \end{array} Using $x^2-y^2=(x+y)(x-y)$ or the factoring of the difference of 2 squares, then the factored form of the expression, $ (3p-5)(9p^2-25) ,$ is \begin{array}{l} (3p-5)(3p+5)(3p-5) \\\\= (3p+5)(3p-5)(3p-5) \\\\= (3p+5)(3p-5)^2 \end{array}