The given mathematical expression is:

$\frac{(-1)^3 + (-\frac{2}{3})^{-2}}{((-4)^{-3})^{-1} - 1}.$

Let us solve it step by step:

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Step 1: Simplify the numerator

The numerator is:
$(-1)^3 + (-\frac{2}{3})^{-2}.$

1. $(-1)^3 = -1$.
2. $(-\frac{2}{3})^{-2}$:
   Recall that for $x^{-n}$, $x^{-n} = \frac{1}{x^n}$. Thus:
   $(-\frac{2}{3})^{-2} = \frac{1}{(-\frac{2}{3})^2} = \frac{1}{\frac{4}{9}} = \frac{9}{4}.$

So, the numerator becomes:
$-1 + \frac{9}{4} = \frac{-4}{4} + \frac{9}{4} = \frac{5}{4}.$

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Step 2: Simplify the denominator

The denominator is:
$((-4)^{-3})^{-1} - 1.$

1. Simplify $(-4)^{-3}$:
   $(-4)^{-3} = \frac{1}{(-4)^3} = \frac{1}{-64} = -\frac{1}{64}.$

2. $(-4)^{-3})^{-1}$:
   Using the rule $(x^{-a})^{-b} = x^{ab}$,
   $(-\frac{1}{64})^{-1} = -64.$

Thus, the denominator becomes:
$-64 - 1 = -65.$

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Step 3: Final computation

The entire expression is:
$\frac{\frac{5}{4}}{-65}.$

Simplify:
$\frac{5}{4} \div -65 = \frac{5}{4 \cdot (-65)} = \frac{5}{-260} = -\frac{1}{52}.$

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Final Answer

$\boxed{-\frac{1}{52}}$

Let me know if you'd like a detailed explanation of any step! 😊

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Here are 5 questions that expand on this topic:

1. How would the answer change if the numerator included a different power, such as $(-1)^5$?
2. Can you prove the general rule for $x^{-n} = \frac{1}{x^n}$?
3. How do negative bases affect even vs. odd exponents?
4. What happens if the denominator changes to include a fraction instead of an integer?
5. How is this problem related to exponent rules in logarithms?

Tip: Always check for negative signs when dealing with powers and fractions—they often lead to common errors!