## College Algebra (11th Edition)

Published by Pearson

# Chapter R - Section R.4 - Factoring Polynomials - R.4 Exercises: 66

#### Answer

(3$z^{3}$ + 4$y^{4}$)(9$z^{6}$ - 12$z^{3}y^{4}$ + 16$y^{8}$)

#### Work Step by Step

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$y^{8}$)

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