Question

1. $5c^{-7}d^{2}cdot(-cd^{2})^{4}$

Explanation:

Step1: Simplify $(-cd^{2})^{4}$

According to power - of - a - product rule $(ab)^n=a^n b^n$, we have $(-cd^{2})^{4}=(-1)^4c^{4}(d^{2})^{4}=c^{4}d^{8}$.

Step2: Multiply the two terms

Multiply $5c^{-7}d^{2}$ and $c^{4}d^{8}$. According to the rule $a^m\times a^n=a^{m + n}$, we get $5c^{-7}d^{2}\times c^{4}d^{8}=5c^{-7 + 4}d^{2+8}$.

Step3: Simplify the exponent of $c$

$c^{-7 + 4}=c^{-3}=\frac{1}{c^{3}}$. So $5c^{-7 + 4}d^{2+8}=\frac{5d^{10}}{c^{3}}$.

Answer:

$\frac{5d^{10}}{c^{3}}$