Basic College Mathematics (10th Edition)

Published by Pearson
ISBN 10:
ISBN 13:

Chapter 3 - Adding and Subtracting Fractions - Mixed Review Exercises - Page 258: 14

Answer

$\frac{11}{27}$

Work Step by Step

$(\frac{2}{3})^{3}+(\frac{2}{3}-\frac{5}{9})$ = $(\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3})+(\frac{2}{3}-\frac{5}{9})$ (First solve bracket) = $(\frac{2\times2\times2}{3\times3\times3})+(\frac{2\times3}{3\times3}-\frac{5}{9})$ (least common multiple of 3 &9 is 9) = $\frac{8}{27}+(\frac{6}{9}-\frac{5}{9})$ = $\frac{8}{27}+(\frac{6-5}{9})$ (subtract numerators) = $\frac{8}{27}+(\frac{1}{9})$ = $\frac{8}{27}+(\frac{1\times3}{9\times3})$ (least common multiple of 9 &27 is 27) = $\frac{8}{27}+(\frac{3}{27})$ = $\frac{8+3}{27}$ = $\frac{11}{27}$
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