Key Concepts and Formulas

This problem requires the application of the Binomial Theorem for expansion of polynomial expressions and strategic use of fundamental algebraic identities to simplify the given product.

1. Binomial Theorem: For any real numbers $a$ and $b$, and any non-negative integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient. A special case often used is for $(1-x)^n$, where the general term is $\binom{n}{k} (1)^{n-k} (-x)^k = \binom{n}{k} (-1)^k x^k$.

2. Algebraic Identities:

   • Difference of Squares: $(a-b)(a+b) = a^2 - b^2$
   • Difference of Cubes: $(a-b)(a^2+ab+b^2) = a^3 - b^3$. In our case, this translates to $(1-x)(1+x+x^2) = 1-x^3$.

Step-by-Step Derivation

Step 1: Simplify the given expression using algebraic identities.

The given expression is $P(x) = (1+x)(1-x)^{10}(1+x+x^2)^9$. Our first goal is to simplify this product into a form that is easier to expand using the Binomial Theorem. We can do this by strategically grouping terms and applying the algebraic identities mentioned above.

First, let's separate one $(1-x)$ term from $(1-x)^{10}$: $P(x) = (1+x)(1-x) \cdot (1-x)^9 \cdot (1+x+x^2)^9$

Now, we can apply the Difference of Squares identity to the first two terms $(1+x)(1-x)$: $(1+x)(1-x) = (1^2 - x^2) = (1-x^2)$ Substituting this back into the expression: $P(x) = (1-x^2) \cdot (1-x)^9 \cdot (1+x+x^2)^9$

Next, we notice that the terms $(1-x)^9$ and $(1+x+x^2)^9$ both have the same power, $9$. This allows us to combine them: $P(x) = (1-x^2) \cdot [(1-x)(1+x+x^2)]^9$ Now, we can apply the Difference of Cubes identity to the expression inside the square brackets, $(1-x)(1+x+x^2)$: $(1-x)(1+x+x^2) = (1^3 - x^3) = (1-x^3)$ Substituting this simplified form back into the expression: $P(x) = (1-x^2)(1-x^3)^9$ This simplified form is much easier to work with. We need to find the coefficient of $x^{18}$ in this product.

Step 2: Expand the $(1-x^3)^9$ term using the Binomial Theorem.

Let's expand the term $(1-x^3)^9$ using the Binomial Theorem. Here, $a=1$, $b=-x^3$, and $n=9$. The general term in the expansion of $(1-x^3)^9$ is given by: $T_{k+1} = \binom{9}{k} (1)^{9-k} (-x^3)^k = \binom{9}{k} (-1)^k (x^3)^k = \binom{9}{k} (-1)^k x^{3k}$ This expansion will only contain terms where the power of $x$ is a multiple of $3$ (i.e., $x^0, x^3, x^6, \dots, x^{27}$).

Step 3: Identify terms contributing to $x^{18}$ in the full product.

The full product is $P(x) = (1-x^2)(1-x^3)^9$. Let the expansion of $(1-x^3)^9$ be $E(x) = A_0 + A_3x^3 + A_6x^6 + \dots + A_{18}x^{18} + \dots$. Then $P(x) = (1-x^2)E(x) = 1 \cdot E(x) - x^2 \cdot E(x)$.

To find the coefficient of $x^{18}$ in $P(x)$, we consider two cases:

• Case 1: The term $1$ from $(1-x^2)$ multiplies a term with $x^{18}$ from $E(x)$. We need the coefficient of $x^{18}$ in $E(x) = (1-x^3)^9$. Using the general term $T_{k+1} = \binom{9}{k} (-1)^k x^{3k}$, we set $3k = 18$, which implies $k=6$. The coefficient for this term is $\binom{9}{6} (-1)^6 = \binom{9}{6}$.

• Case 2: The term $-x^2$ from $(1-x^2)$ multiplies a term with $x^{16}$ from $E(x)$. We need the coefficient of $x^{16}$ in $E(x) = (1-x^3)^9$. Using the general term $T_{k+1} = \binom{9}{k} (-1)^k x^{3k}$, we set $3k = 16$. For $3k=16$, $k = \frac{16}{3}$. Since $k$ must be a non-negative integer for the Binomial Theorem, there is no term with $x^{16}$ in the expansion of $(1-x^3)^9$. Therefore, the coefficient of $x^{16}$ in $E(x)$ is $0$.

Step 4: Calculate the coefficients for each contributing term and sum them.

• From Case 1: The coefficient is $\binom{9}{6}$. $\binom{9}{6} = \binom{9}{9-6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$
• From Case 2: The coefficient is $0$.

Combining these, the total coefficient of $x^{18}$ in the product is $84 + 0 = 84$.

Tips and Common Mistakes

• Prioritize Simplification: Always look for opportunities to simplify complex polynomial products using algebraic identities before attempting binomial expansion. This can significantly reduce the complexity of the problem.
• Binomial Theorem for $(1-x)^n$: Remember the alternating signs. The term is $\binom{n}{k} (-1)^k x^k$.
• Integer Powers: When expanding terms like $(1-x^m)^n$, only powers of $x$ that are multiples of $m$ will appear. If you need a coefficient for a power of $x$ that is not a multiple of $m$, its coefficient is $0$.
• Careful with Distribution: When multiplying two polynomials, ensure you consider all combinations of terms that lead to the desired power of $x$.

Summary

The key to solving this problem efficiently was to first simplify the given product $(1+x)(1-x)^{10}(1+x+x^2)^9$ into $(1-x^2)(1-x^3)^9$ by judiciously applying the difference of squares and difference of cubes identities. Once simplified, we used the Binomial Theorem to expand $(1-x^3)^9$ and then identified the terms that would yield $x^{18}$ when multiplied by $(1-x^2)$. The final coefficient of $x^{18}$ is $84$.