Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set - Page 338: 46

$(a^3-b^4c^5)(a^6+a^3b^4c^5+b^8c^{10})$

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $ a^{9}+b^{12}c^{15} ,$ is \begin{array}{l} (a^3-b^4c^5)[ (a^3)^2-(a^3)(-b^4c^5)+(-b^4c^5)^2] \\\\= (a^3-b^4c^5)(a^6+a^3b^4c^5+b^8c^{10}) .\end{array}