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: 62



Work Step by Step

The expressions $ (x^{2a}-1)^3 $ and $ x^{6a} $ are both perfect cubes (the cube root is exact). Hence, $ (x^{2a}-1)^3-x^{6a} $ 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}-1)^3-(x^{2a})^3 \\\\= [(x^{2a}-1)-x^{2a}][(x^{2a}-1)^2+(x^{2a}-1)(x^{2a})+(x^{2a})^2] \\\\= (x^{2a}-1-x^{2a})[(x^{4a}-2(x^{2a})(1)+(1)^2)+x^{4a}-x^{2a}+x^{4a}] \\\\= -1(x^{4a}-2x^{2a}+1+x^{4a}-x^{2a}+x^{4a}) \\\\= -1(3x^{4a}-3x^{2a}+1) \\\\= -3x^{4a}+3x^{2a}-1 .\end{array}
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