Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.4 Factoring Polynomials - R.4 Exercises - Page 44: 109

Answer

2(y)(3$x^2$ + $y^2$)

Work Step by Step

$(x+y)^3$ - $(x-y)^3$ Now, $a^3$ - $b^3$ = (a-b)($a^2$ + $b^2$ + ab) and a = x+y , b = x-y Which implies, = ((x+y) - (x-y))($(x+y)^2$ + $(x-y)^2$ + (x+y)(x-y)) Now, $(a + b)^2$ = $a^2$ + $b^2$ + 2ab and a = x , b = y Now, $(a - b)^2$ = $a^2$ + $b^2$ - 2ab and a = x , b = y Now, $a^2$ - $b^2$ = (a+b)(a-b) and a = x , b = y Which implies, = 2(y)(($x^2$ + $y^2$ + 2xy) + ($x^2$ + $y^2$ - 2xy) + ($x^2$ - $y^2$)) = 2(y)(3$x^2$ + $y^2$) which is the answer
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