## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set: 79

#### Answer

$-y\sqrt[3]{2x}$

#### Work Step by Step

Simplify each radicand to obtain $\sqrt[3]{(27y^3)(2x)}-y\sqrt[3]{64(2x)} \\=\sqrt[3]{(3y)^3(2x)}-y\sqrt[3]{4^3(2x)} \\=3y\sqrt[3]{2x} - 4y\sqrt [3]{2x}.$ RECALL: The distributive property states that for any real numbers a, b, and c, $ac + bc = (a+b)c \\ac - bc = (a-b)c.$ Use the distributive to obtain $(3y-4y)\sqrt[3]{2x} \\=-y\sqrt[3]{2x}.$

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