College Algebra (11th Edition)

(3$z^{3}$ + 4$y^{4}$)(9$z^{6}$ - 12$z^{3}y^{4}$ + 16$y^{8}$)
27$z^{9}$ + 64$y^{12}$ is a sum of cubes, so we write it as a sum of cubes and factor: $(3z^{3})^{3}$ + $(4y^{4})^{3}$ = (3$z^{3}$ + 4$y^{4}$)($(3z^{3})^{2}$ - 3$z^{3}$( 4$y^{4}$) + $(4y^{4})^{2}$ ) = (3$z^{3}$ + 4$y^{4}$)(9$z^{6}$ - 12$z^{3}y^{4}$ + 16$y^{8}$)