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: 102

Answer

$\displaystyle \frac{1}{3}(x^{2}+4)^{1/3}(11x^{2}+12)=\frac{1}{3}\sqrt[3]{(x^{2}+4)}(11x^{2}+12)$

Work Step by Step

We simplify: $(x^{2}+4)^{4/3}+x\displaystyle *\frac{4}{3}(x^{2}+4)^{1/3}* 2x=(x^{2}+4)^{1/3}((x^{2}+4)^{3/3}+\frac{8}{3}x^{2})=(x^{2}+4)^{1/3}(x^{2}*\frac{3}{3}+4+\frac{8}{3}x^{2})=(x^{2}+4)^{1/3}(\frac{11}{3}x^{2}+4)= \frac{1}{3}(x^{2}+4)^{1/3}(11x^{2}+12)=\frac{1}{3}\sqrt[3]{(x^{2}+4)}(11x^{2}+12)$
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