Question

1. $(-v^{7}w^{-8})^{3}(-v^{-9}w^{6})^{4}$

Explanation:

Step1: Apply power of a product rule

For \((-v^{7}w^{-8})^{3}\), using \((ab)^n = a^n b^n\) and \((a^m)^n=a^{mn}\), we get \((-1)^3(v^{7})^{3}(w^{-8})^{3}=-1\times v^{21}\times w^{-24}=-v^{21}w^{-24}\).
For \((-v^{-9}w^{6})^{4}\), using the same rules, we get \((-1)^4(v^{-9})^{4}(w^{6})^{4}=1\times v^{-36}\times w^{24}=v^{-36}w^{24}\).

Step2: Multiply the two results

Now multiply \(-v^{21}w^{-24}\) and \(v^{-36}w^{24}\). Using \(a^m\times a^n = a^{m + n}\), for \(v\) terms: \(v^{21}\times v^{-36}=v^{21-36}=v^{-15}\), for \(w\) terms: \(w^{-24}\times w^{24}=w^{-24 + 24}=w^{0}=1\) (since \(a^0 = 1\) for \(a
eq0\)). And the coefficient is \(-1\times1=-1\). So the product is \(-1\times v^{-15}\times1=-v^{-15}\). We can also write this as \(-\frac{1}{v^{15}}\) using the rule \(a^{-n}=\frac{1}{a^{n}}\).

Answer:

\(-\frac{1}{v^{15}}\) (or \(-v^{-15}\))