Key Concepts This problem relies on two fundamental mathematical concepts:

1. Difference of Cubes Identity: The algebraic identity $(a-b)(a^2+ab+b^2) = a^3-b^3$ is crucial for simplifying the given expression.
2. Binomial Theorem: For any real numbers $a$ and $b$, and any non-negative integer $n$, the binomial expansion is given by: $(a+b)^n = \sum_{k=0}^{n} {}^{n}C_k a^{n-k} b^k$ where ${}^{n}C_k = \frac{n!}{k!(n-k)!}$ are the binomial coefficients. The general term in the expansion of $(a+b)^n$ is $T_{k+1} = {}^{n}C_k a^{n-k} b^k$.

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Step-by-Step Solution

1. Simplify the Expression using Algebraic Identities The given expression is $(1-x)^{2008}\left(1+x+x^2\right)^{2007}$ Our first step is to simplify this expression, as direct expansion of such high powers is impractical. We can rewrite $(1-x)^{2008}$ as $(1-x) \cdot (1-x)^{2007}$. This allows us to group terms with the same exponent: $(1-x) \cdot (1-x)^{2007} \cdot (1+x+x^2)^{2007}$ Using the property $a^n b^n = (ab)^n$, we can combine the terms with exponent $2007$: $(1-x) \left[ (1-x)(1+x+x^2) \right]^{2007}$ Now, we apply the difference of cubes identity, $(a-b)(a^2+ab+b^2) = a^3-b^3$. Here, $a=1$ and $b=x$. So, $(1-x)(1+x+x^2) = 1^3 - x^3 = 1-x^3$. Substituting this into our expression, we get a much simpler form: $(1-x)(1-x^3)^{2007}$ This simplification is vital because it transforms a complex product into a form suitable for binomial expansion.

2. Expand the Binomial Term $(1-x^3)^{2007}$ Next, we expand $(1-x^3)^{2007}$ using the Binomial Theorem. Let $a=1$, $b=-x^3$, and $n=2007$. The general term of this expansion, $T_{k+1}$, is: $T_{k+1} = {}^{2007}C_k (1)^{2007-k} (-x^3)^k$ $T_{k+1} = {}^{2007}C_k (-1)^k (x^3)^k$ $T_{k+1} = {}^{2007}C_k (-1)^k x^{3k}$ So, the expansion of $(1-x^3)^{2007}$ can be written as: $\sum_{k=0}^{2007} {}^{2007}C_k (-1)^k x^{3k}$ Notice that all powers of $x$ in this expansion are multiples of 3.

3. Determine the General Term of the Full Expansion Now we multiply the expansion from Step 2 by $(1-x)$: $(1-x) \sum_{k=0}^{2007} {}^{2007}C_k (-1)^k x^{3k}$ We distribute $(1-x)$ across the sum: $1 \cdot \left( \sum_{k=0}^{2007} {}^{2007}C_k (-1)^k x^{3k} \right) - x \cdot \left( \sum_{k=0}^{2007} {}^{2007}C_k (-1)^k x^{3k} \right)$ This gives us two types of terms: Type 1: From the first part ($1 \cdot \text{expansion}$), the terms are of the form ${}^{2007}C_k (-1)^k x^{3k}$ The powers of $x$ in these terms are $0, 3, 6, \ldots$, i.e., all are multiples of 3.

Type 2: From the second part ($-x \cdot \text{expansion}$), the terms are of the form $-x \cdot {}^{2007}C_k (-1)^k x^{3k} = {}^{2007}C_k (-1)^{k+1} x^{3k+1}$ The powers of $x$ in these terms are $1, 4, 7, \ldots$, i.e., all are one more than a multiple of 3.

4. Find the Coefficient of $x^{2012}$ We are looking for the coefficient of $x^{2012}$. We need to check if $x^{2012}$ can be generated from either Type 1 or Type 2 terms.

Case A: From Type 1 terms ($x^{3k}$) We need $x^{3k} = x^{2012}$, which implies: $3k = 2012$ $k = \frac{2012}{3}$ To check if $2012$ is divisible by $3$, we sum its digits: $2+0+1+2 = 5$. Since $5$ is not divisible by $3$, $2012$ is not divisible by $3$. Therefore, $k$ is not an integer. The index $k$ in a binomial expansion must be a non-negative integer. Since $k$ is not an integer, there is no term of the form $x^{2012}$ from this part of the expansion.

Case B: From Type 2 terms ($x^{3k+1}$) We need $x^{3k+1} = x^{2012}$, which implies: $3k+1 = 2012$ $3k = 2011$ $k = \frac{2011}{3}$ To check if $2011$ is divisible by $3$, we sum its digits: $2+0+1+1 = 4$. Since $4$ is not divisible by $3$, $2011$ is not divisible by $3$. Therefore, $k$ is not an integer. Again, since $k$ is not an integer, there is no term of the form $x^{2012}$ from this part of the expansion.

Since $x^{2012}$ cannot be formed from either type of term, its coefficient in the expansion is $0$.

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Tips and Common Mistakes to Avoid

• Always simplify first: Before diving into binomial expansion, look for algebraic identities that can simplify the expression. This often significantly reduces complexity. In this problem, recognizing $(1-x)(1+x+x^2) = 1-x^3$ was the key.
• General Term is Powerful: Understand and correctly apply the formula for the general term of a binomial expansion. Pay close attention to signs (like $(-1)^k$).
• Integer Index Requirement: Remember that the index $k$ in the binomial expansion $T_{k+1}$ must be a non-negative integer. If your calculation for $k$ yields a fraction, it means that particular power of $x$ does not exist in the expansion.
• Modular Arithmetic Check: For powers of the form $x^{ak+b}$, checking the desired power modulo $a$ can quickly tell you if a term is possible. Here, we were looking for $x^{2012}$, and the generated terms had powers $3k$ (i.e., $0 \pmod 3$) or $3k+1$ (i.e., $1 \pmod 3$). Since $2012 \equiv 2 \pmod 3$, it's immediately clear that no such term can exist.

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Summary and Key Takeaway By first simplifying the expression using the difference of cubes identity, we transformed the problem into finding the coefficient of $x^{2012}$ in $(1-x)(1-x^3)^{2007}$. Expanding this product revealed that all terms of $x$ would have powers that are either a multiple of 3 (i.e., $3k$) or one more than a multiple of 3 (i.e., $3k+1$). Since $2012$ is two more than a multiple of 3 ($2012 \equiv 2 \pmod 3$), no term with $x^{2012}$ can be formed in the expansion. Therefore, the coefficient of $x^{2012}$ is $0$.

The final answer is $\boxed{0}$.