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

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