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:

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

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