## Intermediate Algebra (12th Edition)

$4\sqrt[3]{4y^2}-19\sqrt[3]{2y}-5$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $(\sqrt[3]{2y}-5)(4\sqrt[3]{2y}+1) ,$ use the special product on multiplying the sum and difference of like terms. $\bf{\text{Solution Details:}}$ Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent \begin{array}{l}\require{cancel} \sqrt[3]{2y}(4\sqrt[3]{2y})+\sqrt[3]{2y}(1)-5(4\sqrt[3]{2y})-5(1) \\\\= 1(4)\sqrt[3]{2y}(\sqrt[3]{2y})+1(1)\sqrt[3]{2y}-5(4)(\sqrt[3]{2y})-5(1) \\\\= 4\sqrt[3]{2y}(\sqrt[3]{2y})+\sqrt[3]{2y}-20\sqrt[3]{2y}-5 .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to \begin{array}{l}\require{cancel} 4\sqrt[3]{2y(2y)}+\sqrt[3]{2y}-20\sqrt[3]{2y}-5 \\\\= 4\sqrt[3]{4y^2}+\sqrt[3]{2y}-20\sqrt[3]{2y}-5 .\end{array} By combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} 4\sqrt[3]{4y^2}+(1-20)\sqrt[3]{2y}-5 \\\\= 4\sqrt[3]{4y^2}-19\sqrt[3]{2y}-5 .\end{array}