SAT Question 2289b199: Equivalent Expressions answer and explanation with Desmos solution

Choice C is correct. For positive values of $a$, $\sqrt[N]{a}=a^{\frac{1}{N}}$. Therefore, the given expression, $\frac{\sqrt[3]{n^{17}{p}^5}}{5n^2\left(\sqrt[8]{p^8}\right)}$, can be rewritten as $\frac{{\left(n^{17}{p}^5\right)}^{\frac{1}{3}}}{5n^2{\left(p^8\right)}^{\frac{1}{8}}}$. Applying properties of exponents, this can be rewritten as $\frac{n^{\frac{17}{3}}{p}^{\frac{5}{3}}}{5n^2{p}^{\frac{8}{8}}}$, or $\frac{n^{\frac{17}{3}}{p}^{\frac{5}{3}}}{5n^{2}p}$. Dividing the numerator and the denominator of this expression by $n$ yields $\frac{n^{\frac{14}{3}}{p}^{\frac{5}{3}}}{5np}$, which is statement I. Dividing the numerator and the denominator of this expression by $np$ yields $\frac{n^{\frac{11}{3}}{p}^{\frac{2}{3}}}{5}$. This expression is equivalent to $\frac{{\left(n^{11}{p}^2\right)}^{\frac{1}{3}}}{5}$. It follows that this expression is equivalent to $\frac{\sqrt[3]{n^{11}{p}^2}}{5}$, which is statement II. Therefore, for all positive values of $n$ and $p$, the given expression is equivalent to I and II.