College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.4 - Factoring Polynomials - R.4 Exercises: 68

Answer

b($b^{2}$ + 9b + 27)

Work Step by Step

$(b + 3)^{3}$ - 27 is a difference of cubes, so we write it as a diference of cubes and factor: $(b + 3)^{3}$ - $3^{3}$ = (b + 3 - 3)($(b + 3)^{2}$ + (b + 3)(3) + $3^{2}$) = b($b^{2}$ + 6b + 9 + 3b + 9 + 9) = b($b^{2}$ + 9b + 27)
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