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

$\sqrt[7]{c-d}$
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt[14]{c^2-2cd+d^2} \\\\= \sqrt[14]{(c-d)^2} .\end{array} \begin{array}{l}\require{cancel} \sqrt[14]{(c-d)^2} \\\\= \left( (c-d)^2 \right)^{1/14} .\end{array} Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \left( (c-d)^2 \right)^{1/14} \\\\= (c-d)^{2\cdot\frac{1}{14}} \\\\= (c-d)^{\frac{1}{7}} .\end{array} Using $x^{m/n}=\sqrt[n]{x^m}=\left(\sqrt[n]{x} \right)^m,$ then \begin{array}{l}\require{cancel} (c-d)^{\frac{1}{7}} \\\\= \sqrt[7]{(c-d)^1} \\\\= \sqrt[7]{c-d} .\end{array}