Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.3 - Algebraic Expressions - 1.3 Exercises - Page 34: 129



Work Step by Step

$(x^{2}+3)^{-1/3}-\dfrac{2}{3}x^{2}(x^{2}+3)^{-4/3}$ Take out common factor $(x^{2}+3)^{-4/3}$: $(x^{2}+3)^{-1/3}-\dfrac{2}{3}x^{2}(x^{2}+3)^{-4/3}=...$ $...=(x^{2}+3)^{-4/3}[(x^{2}+3)-\dfrac{2}{3}x^{2}]=...$ Simplify the expression inside brackets: $...=(x^{2}+3)^{-4/3}\Big[\dfrac{1}{3}x^{2}+3\Big]=...$ Rearrange: $...=\dfrac{\dfrac{1}{3}x^{2}+3}{(x^{2}+3)^{4/3}}=\dfrac{x^{2}+9}{3(x^{2}+3)^{4/3}}=\dfrac{x^{2}+9}{3\sqrt[3]{(x^{2}+3)^{4}}}=...$ $...=\dfrac{x^{2}+9}{3(x^{2}+3)\sqrt[4]{x^{2}+3}}$
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