Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 3 - Exponents, Polynomials and Functions - 3.5 Special Factoring Techniques - 3.5 Exercises: 51

Answer

$(a^{2}+b^{6})(a^4-a^2b^6+b^{12})(a+b^{3})(a^2-ab^3+b^6)(a-b^{3})(a^2+ab^3+b^6)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ a^{12}-b^{36} ,$ use the factoring of the difference of $2$ squares and the factoring of the sum or difference o $2$ cubes. $\bf{\text{Solution Details:}}$ Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (a^{6})^2-(b^{18})^2 \\\\= (a^{6}+b^{18})(a^{6}-b^{18}) \\\\= (a^{6}+b^{18})[(a^{3})^2-(b^{9})^2] \\\\= (a^{6}+b^{18})(a^{3}+b^{9})(a^{3}-b^{9}) .\end{array} 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} [(a^{2})^3+(b^{6})^3](a^{3}+b^{9})(a^{3}-b^{9}) \\\\= (a^{2}+b^{6})(a^4-a^2b^6+b^{12})(a^{3}+b^{9})(a^{3}-b^{9}) \\\\= (a^{2}+b^{6})(a^4-a^2b^6+b^{12})[(a)^{3}+(b^{3})^3](a^{3}-b^{9}) \\\\= (a^{2}+b^{6})(a^4-a^2b^6+b^{12})(a+b^{3})(a^2-ab^3+b^6)(a^{3}-b^{9}) \\\\= (a^{2}+b^{6})(a^4-a^2b^6+b^{12})(a+b^{3})(a^2-ab^3+b^6)[(a)^{3}-(b^{3})^3] \\\\= (a^{2}+b^{6})(a^4-a^2b^6+b^{12})(a+b^{3})(a^2-ab^3+b^6)(a-b^{3})(a^2+ab^3+b^6) .\end{array}
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