College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.6 - Page 86: 22


$\frac{1}{(x^{2}-3x+9)} ; x \ne -3$

Work Step by Step

$\frac{(x^{2}+6x+9)}{(x^{3}+27)}. \frac{1}{(x+3)}$ $(x^{2}+6x+9)$ is in the form of $(a+b)^{2}$ $[(a+b)^{2} = a^{2} + 2ab +b^{2}]$ $[(x+3)^{2} = x^{2} + 2x(3) +3^{2} = x^{2}+6x+9]$ $x^{3}+27$ can be written as $x^{3}+3^{3}$ Factors of $x^{3}+3^{3}$ are $(x+3)(x^{2}-3x+9)$ Because $A^{3}+B^{3}$ = $(A+B)(A^{2}-AB+B^{2})$ So, $\frac{(x^{2}+6x+9)}{(x^{3}+27)}. \frac{1}{(x+3)} $ = $\frac{(x+3)(x+3)}{(x+3)(x^{2}-3x+9)}. \frac{1}{(x+3)}$ If $x=-3$, it makes the denominator 0, so we exclude -3 for $x$. $=\frac{(x+3)(x+3)}{(x+3)(x^{2}-3x+9)}. \frac{1}{(x+3)} ; x\ne-3$ Divide numerator and denominator by common factors. $=\frac{1}{(x^{2}-3x+9)} ; x\ne-3;$
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