## Intermediate Algebra (12th Edition)

$2$
$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $(\sqrt[3]{3}-1)(\sqrt[3]{9}+\sqrt[3]{3}+1) ,$ use the factoring of 2 cubes. $\bf{\text{Solution Details:}}$ Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} (\sqrt[3]{3}-1)[(\sqrt[3]{3})^2+\sqrt[3]{3}(1)+(1)^2] \\\\= (\sqrt[3]{3})^3-(1)^3 \\\\= 3-1 \\\\= 2 .\end{array}