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: 77

Answer

$\color{blue}{(3y^3+5z^2)(9y^6-15y^3z^2+25z^4)}$

Work Step by Step

With $27y^9=(3y^3)^3$ and $125z^6=(5z^2)^3$, the given polynomial is equivalent to: $=(3y^3)^3+(5z^2)^3$ Factor using the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=3y^3$ and $b=5z^2$ to obtain: $=(3y^3+5z^2)[(3y^3)^2-(3y^3)(5z^2) + (5z^2)^2] \\=\color{blue}{(3y^3+5z^2)(9y^6-15y^3z^2+25z^4)}$
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