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: 78

Answer

$\color{blue}{(3z^3+4y^4)(9z^6-12z^3y^4+16y^8)}$

Work Step by Step

With $27z^9=(3z^3)^3$ and $64y^{12}=(4y^4)^3$, the given polynomial is equivalent to: $=(3z^3)^3+(4y^4)^3$ Factor using the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=3z^3$ and $b=4y^4$ to obtain: $=(3z^3+4y^4)[(3z^3)^2-(3z^3)(4y^4) + (4y^4)^2] \\=\color{blue}{(3z^3+4y^4)(9z^6-12z^3y^4+16y^8)}$
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