By definition, 

$\displaystyle b^{-x} = \frac{1}{b^{x}}$.

In our problem, $\displaystyle b = 3$ and $\displaystyle x = 2$. 

Then, we have $\displaystyle \frac{1}{3^{2}} = \frac{1}{9}$.

Raise both sides of the equation to the inverse power of $\displaystyle -3$ to cancel the exponent on the left hand side of the equation.

$\displaystyle \rightarrow ((x+5)^{-3})^{-\frac{1}{3}} = (-1)^{-\frac{1}{3}}$

$\displaystyle \rightarrow x+5 = -1$

Subtract $\displaystyle 5$ from both sides:

$\displaystyle \rightarrow (x+5) - 5 = (-1)-5$

$\displaystyle \rightarrow x = -6$

Remember that exponents in the denominator refer to the root of the term, while exponents in the numerator can be treated normally.

$\displaystyle x^{\frac{a}{b}}=\sqrt[b]{x^a}$

$\displaystyle x^{\frac{3}{7}}=\sqrt[7]{x^3}$

By definition, a number raised to the $\displaystyle \frac{1}{2}$ power is the same as the square root of that number. 

Since the square root of 64 is 8, 8 is our solution. 

Remember that fraction exponents are the same as radicals.

$\displaystyle \small 16^{\frac{1}{2}}=\sqrt{16}=4$

$\displaystyle 256^{\frac{3}{4}}=\sqrt[4]{256^3}=64$

A shortcut would be to express the terms as exponents and look for opportunities to cancel.

$\displaystyle 16^{\frac{1}{2}}=(4^2)^{\frac{1}{2}}=4$

$\displaystyle \small 256^{\frac{3}{4}}=(4^4)^{\frac{3}{4}}=4^3=64$

Either method, we then need to multiply to two terms.

$\displaystyle \small (4)(64)=256$

When dividing with exponents, the exponent in the denominator is subtracted from the exponent in the numerator. For example: $\displaystyle \frac{x^3}{x^6}= x^{3-6}=x^{-3}=\frac{1}{x^3}$.

In our problem, each term can be treated in this manner. Remember that a negative exponent can be moved to the denominator.

$\displaystyle \frac{4x^3y^8z^2}{8x^6y^2z^4}=\frac{4}{8}x^{-3}y^6z^{-2}=\frac{4y^6}{8x^3z^2}$

Now, simplifly the numerals.

$\displaystyle \frac{4y^6}{8x^3z^2}=\frac{y^6}{2x^3z^2}$

We are given: $\displaystyle 2^{3} \cdot 2^{6}$. 

Recall that when we are multiplying exponents with the same base, we keep the base the same and add the exponents. 

Thus, we have $\displaystyle 2^{3 + 6} = 2^{9}$.

Recall that when we are dividing exponents with the same base, we keep the base the same and subtract the exponents. 

Thus, we have $\displaystyle \frac{2^{9}}{2^{11}} = 2^{9 - 11} = 2^{-2}$.

We also recall that for negative exponents,

$\displaystyle b^{-x} = \frac{1}{b^{x}}$.

Thus, $\displaystyle 2^{-2} = \frac{1}{2^{2}}$.

Begin by rearranging the terms in the numerator and denominator so that the exponents are positive:

$\displaystyle (\frac{a^{\frac{-3}{2}}}{3^6b^{\frac{-2}{3}}})^{\frac{-1}{2}}$

$\displaystyle (\frac{3^6b^{\frac{-2}{3}}}{a^{\frac{-3}{2}}})^{\frac{1}{2}}$

$\displaystyle (\frac{3^6a^{\frac{3}{2}}}{b^{\frac{2}{3}}})^{\frac{1}{2}}$

Multiply the exponents:

$\displaystyle (\frac{3^3a^{\frac{3}{4}}}{b^{\frac{1}{3}}})$

Simplify:

$\displaystyle \frac{27a^{\frac{3}{4}}}{b^{\frac{1}{3}}}$

First simplify the second term, and then combine the two:

$\displaystyle x^3 + \frac{2x^2}{x^{-1}}$

$\displaystyle = x^3 + 2x^{2-(-1)} = x^3 + 2x^3 = 3x^3$