Answer
$3y^2*(x+3)(x^2-3x+9)$
Work Step by Step
$3x^6y^2+81y^2$
$y^2*(3x^6+81)$
$3y^2*(x^6+27)$
$x^6+27=x^3+3^3$
$a^3+b^3=(a+b)(a^2-ab+b^2)$
$x^6+27= (x+3)(x^2-3x+3^2)$
$x^6+27= (x+3)(x^2-3x+9)$
$3y^2*(x^6+27)$
$3y^2*(x+3)(x^2-3x+9)$
Algebra: A Combined Approach (4th Edition)
by Martin-Gay, Elayn
Published by
Pearson
ISBN 10:
0321726391
ISBN 13:
978-0-32172-639-1
Chapter 6 - Section 6.5 - Factoring by Special Products - Exercise Set - Page 448: 77
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