College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.8 - nth Roots; Rational Exponents - R.8 Assess Your Understanding - Page 80: 108

Answer

$12(6x+1)^{1/3}(4x-3)^{1/2}(5x-1)$

Work Step by Step

We simplify: $6(6x+1)^{1/3}(4x-3)^{3/2}+6(6x+1)^{4/3}(4x-3)^{1/2} =6(6x+1)^{1/3}(4x-3)^{1/2}[(4x-3)^{2/2}+(6x+1)^{3/3}] =6(6x+1)^{1/3}(4x-3)^{1/2}[(4x-3)+(6x+1)] =6(6x+1)^{1/3}(4x-3)^{1/2}(10x-2)\\ =6*2(6x+1)^{1/3}(4x-3)^{1/2}(5x-1)\\ =12(6x+1)^{1/3}(4x-3)^{1/2}(5x-1)$
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