Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\sqrt[3]{x^2+xy+y^2}$
Using the properties of radicals, the given expression, $\dfrac{\sqrt[3]{x^3-y^3}}{\sqrt[3]{x-y}} ,$ simplifies to \begin{array}{l}\require{cancel} \sqrt[3]{\dfrac{x^3-y^3}{x-y}} \\\\= \sqrt[3]{\dfrac{(x-y)(x^2+xy+y^2)}{x-y}} \\\\= \sqrt[3]{\dfrac{(\cancel{x-y})(x^2+xy+y^2)}{\cancel{x-y}}} \\\\= \sqrt[3]{x^2+xy+y^2} .\end{array} * Note that it is assumed that all variables represent positive numbers.