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 4 - Polynomials - 4.6 Special Products - 4.6 Exercise Set - Page 278: 111



Work Step by Step

RECALL: $x^my^m=(xy)^m$ Use the rule above to obtain: $=[(t^3-1)(t^3+1)]^4 \\=(t^3-1)(t^3+1) \cdot (t^3-1)(t^3+1) \cdot (t^3-1)(t^3+1) \cdot (t^3-1)(t^3+1)$ Use the formula $(a-b)(a+b) = a^2-b^2$ with $a=t^3$ and $b=1$ to obtain: $=[(t^3)^2-1^2] \cdot [(t^3)^2-1^2] \cdot [(t^3)^2-1^2] \cdot [(t^3)^2-1^2] \\=(t^6-1) \cdot (t^6-1) \cdot (t^6-1) \cdot (t^6-1) \\=[(t^6-1) \cdot (t^6-1)] \cdot [(t^6-1) \cdot (t^6-1)]$ Use the formula $(a-b)(a-b) = a^2-2ab+b^2$ with $a=t^6$ and $b=1$ to obtain: $=[(t^6)^2-2(t^6)(1)+1^2] \cdot [(t^6)^2-2(t^6)(1)+1^2] \\=(t^{12}-2t^6+1) \cdot (t^{12}-2t^6+1)$ Distribute each term of the first trinomial to the second trinomial to obtain: $=t^{12}(t^{12}-2t^6+1)-2t^6(t^{12}-2t^6+1)+1(t^{12}-2t^6+1) \\=t^{12}(t^{12}) -t^{12}2t^6+t^{12}(1)-2t^6(t^{12})-2t^6(-2t^6)-2t^6(1)+(t^{12}-2t^6+1) \\=t^{24}-2t^{18}+t^{12}-2t^{18}+4t^{12}-2t^6+t^{12}-2t^6+1$ Combine like terms to obtain: $\\=t^{24}+(-2t^{18}-2t^{18})+(t^{12}+4t^{12}+t^{12})+(-2t^6-2t^6)+1 \\=t^{24}+(-4t^{18})+6t^{12}+(-4t^6)+1 \\=t^{24}-4t^{18}+6t^{12}-4t^6+1$
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