Question

1. \\(\left(\frac{2v^3 \cdot 2v^{-4} \cdot v^{-4}}{v^0}\

ight)^3\\)

Explanation:

Step1: Simplify numerator's coefficients and exponents

First, multiply the coefficients: \(2 \cdot 2 = 4\). Then, for the variable \(v\), use the rule \(a^m \cdot a^n = a^{m + n}\). So \(v^3 \cdot v^{-4} \cdot v^{-4}=v^{3 + (-4) + (-4)} = v^{-5}\). The numerator becomes \(4v^{-5}\), and the denominator is \(v^0 = 1\) (since any non - zero number to the power of 0 is 1). So the fraction inside the parentheses is \(4v^{-5}\).

Step2: Apply the power of a power rule

Now, we have \((4v^{-5})^3\). Using the rule \((ab)^n=a^n b^n\) and \((a^m)^n=a^{m\cdot n}\), we get \(4^3\cdot(v^{-5})^3\). Calculate \(4^3 = 64\) and \((v^{-5})^3=v^{-15}\). So the expression simplifies to \(64v^{-15}\), and we can also write this as \(\frac{64}{v^{15}}\) using the rule \(a^{-n}=\frac{1}{a^n}\).

Answer:

\(64v^{-15}\) (or \(\frac{64}{v^{15}}\))