To find the constant term of $(x - 2x^{-2})^9$, we can use the binomial theorem.

The binomial expansion for $(a + b)^n$ is given by:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

In your case, $a = x$, $b = -2x^{-2}$, and $n = 9$. So, the binomial expansion is:

$(x - 2x^{-2})^9 = \sum_{k=0}^{9} \binom{9}{k} x^{9-k} (-2x^{-2})^k$

Simplifying the powers of $x$ for each term:

$\binom{9}{k} x^{9-k} (-2)^k (x^{-2})^k = \binom{9}{k} (-2)^k x^{9-k-2k} = \binom{9}{k} (-2)^k x^{9-3k}$

We want the constant term, which occurs when the exponent of $x$ is 0. Therefore, we set:

$9 - 3k = 0$

Solving for $k$:

$3k = 9 \quad \Rightarrow \quad k = 3$

Now, substitute $k = 3$ into the binomial expansion formula:

$\text{Term} = \binom{9}{3} (-2)^3 x^{9 - 3(3)} = \binom{9}{3} (-8) x^0$

Since $x^0 = 1$, the constant term is:

$\binom{9}{3} (-8) = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times (-8) = 84 \times (-8) = -672$

So, the constant term is $-672$.

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Would you like a further explanation or have any questions about this? Here are a few related questions you can explore:

1. How do binomial expansions work when exponents are negative?
2. What is the significance of the binomial theorem in combinatorics?
3. Can this method be applied to other types of expressions like $(x + 3x^{-3})^7$?
4. How does the general term formula in a binomial expansion change for higher powers?
5. Can the constant term of an expansion be calculated directly without expanding the entire series?

Tip: When seeking the constant term, always solve for the value of $k$ that makes the exponent of $x$ equal to 0.