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: 52

Answer

$(x^{2}+y^{6})(x^4-x^2y^6+y^{12})(x+y^{3})(x^2-xy^3+y^6)(x-y^{3})(x^2+xy^3+y^6)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ x^{12}-y^{36} ,$ use the factoring of the difference of $2$ squares and the factoring of the sum or difference of $2$ 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} (x^{6})^2-(y^{18})^2 \\\\= (x^{6}+y^{18})(x^{6}-y^{18}) \\\\= (x^{6}+y^{18})[(x^{3})^2-(y^{9})^2] \\\\= (x^{6}+y^{18})(x^{3}+y^{9})(x^{3}-y^{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} [(x^{2})^3+(y^{6})^3](x^{3}+y^{9})(x^{3}-y^{9}) \\\\= (x^{2}+y^{6})(x^4-x^2y^6+y^{12})(x^{3}+y^{9})(x^{3}-y^{9}) \\\\= (x^{2}+y^{6})(x^4-x^2y^6+y^{12})[(x)^{3}+(y^{3})^3](x^{3}-y^{9}) \\\\= (x^{2}+y^{6})(x^4-x^2y^6+y^{12})(x+y^{3})(x^2-xy^3+y^6)(x^{3}-y^{9}) \\\\= (x^{2}+y^{6})(x^4-x^2y^6+y^{12})(x+y^{3})(x^2-xy^3+y^6)[(x)^{3}-(y^{3})^3] \\\\= (x^{2}+y^{6})(x^4-x^2y^6+y^{12})(x+y^{3})(x^2-xy^3+y^6)(x-y^{3})(x^2+xy^3+y^6) .\end{array}
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