The problem asks us to simplify several expressions involving exponents and negative numbers. We will go through each of the problems 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.

1. Problem Description

The problem asks us to simplify several expressions involving exponents and negative numbers. We will go through each of the problems 1, 2, 3, 4, 5, 6, 7, 8, 9, and

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0.

2. Solution Steps

1. $(-3 \cdot (-6))^{-3}$

First simplify the term inside the parentheses: $-3 \cdot (-6) = 18$.

Then we have $18^{-3}$. Using the property $a^{-n} = \frac{1}{a^n}$, we get $18^{-3} = \frac{1}{18^3}$.

$18^3 = 18 \cdot 18 \cdot 18 = 324 \cdot 18 = 5832$.

So, the final expression becomes $\frac{1}{5832}$.

2. $(\frac{-2}{-5})^{-2}$

Since $\frac{-2}{-5} = \frac{2}{5}$, we have $(\frac{2}{5})^{-2}$.

Using the property $(\frac{a}{b})^{-n} = (\frac{b}{a})^n$, we get $(\frac{2}{5})^{-2} = (\frac{5}{2})^2$.

$(\frac{5}{2})^2 = \frac{5^2}{2^2} = \frac{25}{4}$.

3. $\frac{6^4}{6^{-3}}$

Using the quotient rule $\frac{a^m}{a^n} = a^{m-n}$, we have $\frac{6^4}{6^{-3}} = 6^{4 - (-3)} = 6^{4+3} = 6^7$.

$6^7 = 6^3 \cdot 6^4 = 216 \cdot 1296 = 279936$.

4. $\frac{6^4}{6^7}$

Using the quotient rule $\frac{a^m}{a^n} = a^{m-n}$, we have $\frac{6^4}{6^7} = 6^{4-7} = 6^{-3}$.

$6^{-3} = \frac{1}{6^3} = \frac{1}{216}$.

5. $\frac{-3^7}{-3^5}$

$\frac{-3^7}{-3^5} = \frac{-1 \cdot 3^7}{-1 \cdot 3^5} = \frac{3^7}{3^5}$.

Using the quotient rule $\frac{a^m}{a^n} = a^{m-n}$, we get $\frac{3^7}{3^5} = 3^{7-5} = 3^2 = 9$.

6. $-4^{8-5}$

$-4^{8-5} = -4^3 = - (4^3) = - (4 \cdot 4 \cdot 4) = -64$.

7. $-8^{38}$

Since the negative sign is not inside the parenthesis, the expression is $-(8^{38})$. We can leave it as $-8^{38}$.

8. $(-2 \cdot (-4))^{-7}$

First simplify the term inside the parentheses: $-2 \cdot (-4) = 8$.

Then we have $8^{-7}$. Using the property $a^{-n} = \frac{1}{a^n}$, we get $8^{-7} = \frac{1}{8^7}$.

$8^7 = 2097152$. So, the final expression becomes $\frac{1}{2097152}$.

9. $\frac{7^{-9}}{7^{-5}}$

Using the quotient rule $\frac{a^m}{a^n} = a^{m-n}$, we have $\frac{7^{-9}}{7^{-5}} = 7^{-9 - (-5)} = 7^{-9+5} = 7^{-4}$.

$7^{-4} = \frac{1}{7^4} = \frac{1}{2401}$.

1

0. $3^{-4} \cdot 3^{-8}$

Using the product rule $a^m \cdot a^n = a^{m+n}$, we have $3^{-4} \cdot 3^{-8} = 3^{-4 + (-8)} = 3^{-12}$.

$3^{-12} = \frac{1}{3^{12}} = \frac{1}{531441}$.

3. Final Answer

1. $\frac{1}{5832}$

2. $\frac{25}{4}$

3. $279936$

4. $\frac{1}{216}$

5. $9$

6. $-64$

7. $-8^{38}$

8. $\frac{1}{2097152}$

9. $\frac{1}{2401}$

1

0. $\frac{1}{531441}$