Apply the power of a product property:

\(\displaystyle \left (6x ^{-5 } \right )^{-3}\)

\(\displaystyle = 6 ^{-3}\cdot \left (x ^{-5 } \right )^{-3}\)

\(\displaystyle =\frac{1}{216}\cdot x ^{-5 \cdot (-3) }\)

\(\displaystyle =\frac{1}{216}\cdot x ^{15 }\)

\(\displaystyle =\frac{x ^{15 }}{216}\)

By the Binomial Theorem, if \(\displaystyle (x + 1)^{n}\) is expanded, the coefficient of \(\displaystyle x^{r}\) is

 \(\displaystyle _{n}\textrm{C} _{r} = \frac{n!}{(n-r)!r!}\).

Substitute \(\displaystyle n = 12, r = 5\): The coefficient of \(\displaystyle x^{5}\) is:

\(\displaystyle _{12}\textrm{C} _{5} = \frac{12!}{(12-5)!5!} = \frac{12!}{7!5!}\)

\(\displaystyle = \frac{12 \cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}\)

\(\displaystyle = \frac{12 \cdot 11\cdot 10\cdot 9\cdot 8}{ 5\cdot 4\cdot 3\cdot 2\cdot 1} = \frac{11\cdot 9\cdot 8}{ 1} = 792\)

Apply the power of a power property twice:

\(\displaystyle \left [\left ( x ^{3} \right )^{3} \right ]^{3 } = ( x ^{3\; \cdot \; 3 } )^{3 } = ( x ^{9 } )^{3 } = x ^{9\cdot 3} =x ^{27}\)

By the Binomial Theorem, the \(\displaystyle x^{r}\) term of \(\displaystyle (ax + b)^{n}\) is 

\(\displaystyle _{n}\textrm{C} _{r} b^{n-r}\left ( ax \right )^{r}= \frac{n!}{(n-r)!r!} a^{r} b^{n-r} x^{r}\),

making the coefficient of \(\displaystyle x^{r}\)

\(\displaystyle \frac{n!}{(n-r)!r!} a^{r} b^{n-r}\).

We can set \(\displaystyle a = 2,b=3,n=6,r=4\) in this expression:

\(\displaystyle \frac{6!}{(6-4)!4!} \cdot 2^{4} \cdot 3^{6-4}\)

\(\displaystyle =\frac{6!}{2!4!}\cdot 2^{4}\cdot 3^{2}\)

\(\displaystyle =\frac{720}{2\cdot 24}\cdot 16\cdot 9\)

\(\displaystyle =2,160\)

By the Binomial Theorem, the \(\displaystyle x^{r}\) term of \(\displaystyle (x + b)^{n}\) is 

\(\displaystyle _{n}\textrm{C} _{r} b^{n-r} x^{r}= \frac{n!}{(n-r)!r!} b^{n-r} x^{r}\).

Substitute \(\displaystyle b = 3, n = 10, r= 8\) and this becomes

\(\displaystyle \frac{10!}{(10-8)!8!} 3^{10-8} x^{8}\).

The coefficient is 

\(\displaystyle \frac{10!}{(10-8)!8!} \cdot 3^{10-8} = \frac{10!}{2!8!}\cdot 3^{2} = \frac{9 \cdot 10}{2}\cdot 9 = 405\).

We need to apply the power of power rule twice:

\(\displaystyle \left [ (x^4)^4 \right ]^5=(x^{4\times 4})^5=(x^{16})^5=x^{16\times 5}=x^{80}\)

Based on the power of a product rule we have:

\(\displaystyle (x^{-4})^{a}=x^{8}\Rightarrow x^{-4\times a}=x^8\)

The bases are the same, so we can write:

\(\displaystyle -4a=8\Rightarrow a=\frac{8}{-4}=-2\)

First, recognize that raising the fraction to a negative power is the same as raising the inverted fraction to a positive power.

\(\displaystyle (\frac{2x^5}{3y^2})^{-4}=(\frac{3y^2}{2x^5})^4\)

Apply the exponent within the parentheses and simplify.

\(\displaystyle \frac{(3y^2)^4}{(2x^5)^4}\)

\(\displaystyle \frac{3^4(y^2)^4}{2^4(x^5)^4}\)

\(\displaystyle \frac{81y^{(2\times 4)}}{16x^{(5\times 4)}}\)

\(\displaystyle \frac{81y^8}{16x^{20}}\)

This fraction cannot be simplified further.

First, recognize that raising the fraction to a negative power is the same as raising the inverted fraction to a positive power.

\(\displaystyle (\frac{x^2y}{2z^3})^{-2}=(\frac{2z^3}{x^2y})^2\)

Apply the exponent within the parentheses and simplify.

\(\displaystyle \frac{(2z^3)^2}{(x^2y)^2}\)

\(\displaystyle \frac{2^2(z^3)^2}{(x^2)^2y^2}\)

\(\displaystyle \frac{4z^{(3\times 2)}}{x^{(2\times 2)}y^2}\)

\(\displaystyle \frac{4z^6}{x^4y^2}\)

Begin by factoring the numerator and denominator. \(\displaystyle xy\) can be factored out of each term.

\(\displaystyle \frac{x^3y+2x^2y+xy}{xy(x+1)}=\frac{xy(x^2+2x+1)}{xy(x+1)}\)

\(\displaystyle xy\) can be canceled, since it appears in both the numerator and denomintor.

\(\displaystyle \frac{x^2+2x+1}{x+1}\)

Next, factor the numerator.

\(\displaystyle \frac{(x^2+2x+1)}{x+1}=\frac{(x+1)(x+1)}{x+1}\)

Simplify.

\(\displaystyle \frac{(x+1)(x+1)}{x+1}=x+1\)