Answer
$\sqrt[7]{c-d}$
Work Step by Step
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}
