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 79: 90


$\displaystyle \frac{4x+3}{3(x+1)^{2/3}}$

Work Step by Step

$(x+1)^{1/3}+x\displaystyle \cdot\frac{1}{3}(x+1)^{-2/3}=$ ... rewrite the second term moving $(x+1)$ into the denominator $=(x+1)^{1/3}+\displaystyle \frac{x}{3(x+1)^{2/3}}$ ... rewrite, with the first term being a fraction with the common denominator $=\displaystyle \frac{(x+1)^{1/3}\cdot 3(x+1)^{2/3}+x}{3(x+1)^{2/3}}$ ... apply $a^{m}\cdot a^{n}=a^{m+n}$ to the first term of the numerator $=\displaystyle \frac{3(x+1)^{1/3+2/3}+x}{3(x+1)^{2/3}}$ $=\displaystyle \frac{3(x+1)^{1}+x}{3(x+1)^{2/3}}$ $=\displaystyle \frac{3x+3+x}{3(x+1)^{2/3}}$ $=\displaystyle \frac{4x+3}{3(x+1)^{2/3}}$
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