College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 87: 79

Answer

$ \frac{2d}{a^{2}+ab+b^{2}}$

Work Step by Step

$(\frac{1}{a^{3}-b^{3}} . \frac{ac+ad-bc-bd}{1}) - \frac{c-d}{a^{2}+ab+b^{2}}$ $= (\frac{ac+ad-bc-bd}{a^{3}-b^{3}}) - \frac{c-d}{a^{2}+ab+b^{2}}$ $= (\frac{a(c+d)-b(c+d)}{a^{3}-b^{3}}) - \frac{c-d}{a^{2}+ab+b^{2}}$ $= (\frac{(a-b)(c+d)}{a^{3}-b^{3}}) - \frac{c-d}{a^{2}+ab+b^{2}}$ Factors of $a^{3}-b^{3}$ are $(a-b)(a^{2}+ab+b^{2})$ $= (\frac{(a-b)(c+d)}{(a-b)(a^{2}+ab+b^{2})}) - \frac{c-d}{a^{2}+ab+b^{2}}$ $= \frac{c+d}{a^{2}+ab+b^{2}} - \frac{c-d}{a^{2}+ab+b^{2}}$ $= \frac{c+d-c+d}{a^{2}+ab+b^{2}} $ $= \frac{2d}{a^{2}+ab+b^{2}} $
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