In order to simplify this expression, we will need to rewrite the negative exponents.

The negative exponents can be rewritten into fractional form.

\(\displaystyle x^{-b} =\frac{1}{x^b}\)

\(\displaystyle 8^{-2}-2^{-5} = \frac{1}{8^2}-\frac{1}{2^5} = \frac{1}{64}- \frac{1}{32}\)

Convert the second fraction to match the denominator of the first fraction.

\(\displaystyle \frac{1}{64}- \frac{1(2)}{32(2)} = \frac{1}{64}-\frac{2}{64}= -\frac{1}{64}\)

The answer is:  \(\displaystyle -\frac{1}{64}\)

Rewrite the negative exponent as the reciprocal of the positive power.

The negative exponent can be rewritten into a fraction as follows:

\(\displaystyle x^{-a} =\frac{1}{x^a}\)

\(\displaystyle (\frac{2}{3})^{-3}= \frac{1}{(\frac{2}{3})^3} = \frac{1}{\frac{8}{27}}\)

Simplify the complex fraction.

\(\displaystyle \frac{1}{\frac{8}{27}} = 1\div \frac{8}{27} = 1\times \frac{27}{8}= \frac{27}{8}\)

The answer is:  \(\displaystyle \frac{27}{8}\)

The negative exponents can be rewritten into a fraction.

\(\displaystyle x^{-a} =\frac{1}{x^a}\)

Rewrite both terms given in the problem.

\(\displaystyle -4^{-3}-2^{-3}=-\frac{1}{4^3}-\frac{1}{2^3} = -\frac{1}{64}-\frac{1}{8}\)

Find the least common denominator of both fractions.

\(\displaystyle -\frac{1}{64}-\frac{1}{8} = -\frac{1}{64}-\frac{1(8)}{8(8)}\)

Simplify the fractions.

\(\displaystyle -\frac{1}{64}-\frac{8}{64}=-\frac{9}{64}\)

The answer is:  \(\displaystyle -\frac{9}{64}\)

Simplify the negative exponent as follows:

\(\displaystyle x^{-a} =\frac{1}{x^a}\)

\(\displaystyle 8^{-2} = \frac{1}{8^2} = \frac{1}{64}\)

Rewrite the expression.

\(\displaystyle -(-8^{-2})+8 = -(-\frac{1}{64})+8 = \frac{1}{64}+8\)

The answer is:  \(\displaystyle 8\frac{1}{64}\)

Change all negative exponents into a fractional form by the following property.

\(\displaystyle x^{-a} =\frac{1}{x^a}\)

\(\displaystyle 9^{-2}\times\frac{3}{ 2^{-4}} = \frac{1}{9^2} \times \frac{3}{\frac{1}{2^4}} = \frac{1}{81} \times \frac{3}{\frac{1}{16}}\)

Simplify the fractions.

\(\displaystyle \frac{1}{81} \times \frac{3}{\frac{1}{16}} = \frac{1}{81} \times 48 = \frac{48}{81}\)

\(\displaystyle \frac{48}{81} =\frac{16 \times 3}{27 \times 3}\)

The answer is:   \(\displaystyle \frac{16}{27}\)

When dealing with negative exponents, we write \(\displaystyle x^{-a}=\frac{1}{x^a}\). Therefore \(\displaystyle 7^{-3}=\frac{1}{7^3}\).

When dealing with negative exponents, we write \(\displaystyle x^{-a}=\frac{1}{x^a}\). Therefore \(\displaystyle (x+5)^{-3}=\frac{1}{(x+5)^3}\)

When dealing with negative exponents, we write \(\displaystyle x^{-a}=\frac{1}{x^a}\). Therefore \(\displaystyle (\frac{2}{3})^{-2}=\frac{1}{(\frac{2}{3})^2}=\frac{1}{\frac{4}{9}}=\frac{9}{4}\).

When dealing with negative exponents, we write \(\displaystyle x^{-a}=\frac{1}{x^a}\). Therefore \(\displaystyle (\frac{x}{6})^{-3}=\frac{1}{(\frac{x}{6})^3}=\frac{1}{\frac{x^3}{256}}=\frac{256}{x^3}\).

When dealing with exponents, we convert as such: \(\displaystyle x^{-a}=\frac{1}{x^a}\). Therefore, \(\displaystyle 3^{-4}=\frac{1}{3^4}=\frac{1}{81}\).