Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set - Page 339: 61



Work Step by Step

The expressions $ x^{6a} $ and $ (x^{2a}+1)^3 $ are both perfect cubes (the cube root is exact). Hence, $ x^{6a}-(x^{2a}+1)^3 $ is a $\text{ difference }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x^{2a})^3-(x^{2a}+1)^3 \\\\= [x^{2a}-(x^{2a}+1)][(x^{2a})^2+x^{2a}(x^{2a}+1)+(x^{2a}+1)^2] \\\\= [x^{2a}-x^{2a}-1][x^{4a}+x^{4a}+x^{2a}+(x^{4a}+2(x^{4a})(1)+(1)^2] \\\\= [x^{2a}-x^{2a}-1](x^{4a}+x^{4a}+x^{2a}+x^{4a}+2x^{4a}+1) \\\\= -1(5x^{4a}+x^{2a}+1) \\\\= -5x^{4a}-x^{2a}-1 .\end{array}
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