Question

simplify the expression, and then complete the sentence.$sqrt8{\frac{n^{18}p^{7}}{n^{2}p^{-1}}} = square$the restrictions on $n$ and $p$ are such that $n
eq square$ and $p
eq square$.

Explanation:

Step1: Simplify inside the radical

Subtract exponents of like bases:
$\frac{n^{18}p^{7}}{n^{2}p^{-1}} = n^{18-2}p^{7-(-1)} = n^{16}p^{8}$

Step2: Apply 8th root to the result

Take the 8th root of each term:
$\sqrt[8]{n^{16}p^{8}} = (n^{16})^{\frac{1}{8}}(p^{8})^{\frac{1}{8}} = n^{2}|p|$

Step3: Identify variable restrictions

Radical requires non-negative input, denominator can't be zero:
For the original expression, $n^2p^{-1}
eq 0$ so $n
eq 0$ and $p
eq 0$. Also, the radicand must be non-negative; since $n^{16}p^8$ is always non-negative for real $n,p
eq 0$, the only restrictions are $n
eq 0$, $p
eq 0$.

Answer:

$\sqrt[8]{\frac{n^{18}p^{7}}{n^{2}p^{-1}}} = n^2|p|$
The restrictions on $n$ and $p$ are such that $n
eq 0$ and $p
eq 0$.