Answer:
Let's take this problem step-by-step.
1. We should first note the Power of a Product Rule:
[tex](ab)^{x}=a^{x}b^{x}[/tex]
[tex](64n^{12})^{-\frac{1}{6} }=(64^{-\frac{1}{6}})(n^{12})^{-\frac{1}{6} }[/tex]
Since that's a bit hard to read, here's some clarification: we have two parts that are multiplying together. We have:
[tex]64^{-\frac{1}{6}}[/tex] times the term [tex](n^{12})^{-\frac{1}{6} }[/tex]
2. Next, we should also note the Power Rule for exponents, that is:
[tex](x^{m})^{n} =x^{(m)(n)}[/tex]
[tex](n^{12})^{-\frac{1}{6} }= n^{(12)(-\frac{1}{6})}=n^{-2}[/tex]
From our previous equation:
[tex](64^{-\frac{1}{6}})(n^{12})^{-\frac{1}{6} }[/tex]
[tex](64^{-\frac{1}{6}})(n^{-2})[/tex]
3. We should know that, in general:
[tex]x^{\frac{1}{n}}=\sqrt[n]{x}[/tex]
The variable "n" represents any numerical constant. In other words, exponents which are fractions translate to square roots, cubic roots, and so on depending on the denominator. This is the Fractional Exponent Rule.
Therefore,
[tex]x^{\frac{1}{6}}=\sqrt[6]{x}[/tex]
[tex]64^{\frac{1}{6}}=\sqrt[6]{64}[/tex]
[tex]\sqrt[6]{64}=2[/tex]
[tex](64^{-\frac{1}{6}})(n^{-2})=(2^{-1})(n^{-2})[/tex]
4. However, we have to take into account that the exponent is negative. In general:
[tex]n^{-1}=\frac{1}{n}[/tex]
If an exponent is negative, we have to take the reciprocal of the term of the base "n." In simpler terms, we put whatever is attached to the exponent as a fraction with 1 as the numerator. This is the Negative Exponent Rule.
Therefore,
[tex](2^{-1})(n^{-2})=(\frac{1}{2})(\frac{1}{n^{2}} )=\frac{1}{2n^{2}}[/tex]
Our answer is 1/2n^2.
TL;DR
Here are all the necessary equations to solve the problem:
[tex](64n^{12})^{-\frac{1}{6} }=(64^{-\frac{1}{6}})(n^{12})^{-\frac{1}{6} }[/tex]
[tex](64^{-\frac{1}{6}})(n^{12})^{-\frac{1}{6} }=(64^{-\frac{1}{6}})(n^{-2})[/tex]
[tex](64^{-\frac{1}{6}})(n^{-2})=(2^{-1})(n^{-2})[/tex]
[tex](2^{-1})(n^{-2})=(\frac{1}{2})(\frac{1}{n^{2}} )=\frac{1}{2n^{2}}[/tex]
