1. Key Concepts and Formulae

This problem primarily utilizes the Binomial Theorem and fundamental algebraic manipulation. The Binomial Theorem states that for any non-negative integer $n$: $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ A crucial special case is for the expansion of $(1+x)^n$, where the coefficient of $x^k$ is given by $\binom{n}{k}$. We will also employ algebraic identities such as the cube of a sum $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and the difference of squares $a^2 - b^2 = (a-b)(a+b)$.

2. Step-by-Step Working with Clear Explanations

Let the given expression be $E$. $E = (1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5$

Step 1: Simplify the third term using algebraic identity.

Observe the expression inside the third parenthesis: $\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)$. This expression is a perfect cube, specifically $(1+a)^3$ where $a = \frac{1}{x}$. Recall the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. Letting $a=1$ and $b=\frac{1}{x}$, we have: $(1+\frac{1}{x})^3 = 1^3 + 3(1)^2(\frac{1}{x}) + 3(1)(\frac{1}{x})^2 + (\frac{1}{x})^3$ $(1+\frac{1}{x})^3 = 1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3}$ Therefore, we can simplify the third term: $\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5 = \left(\left(1+\frac{1}{x}\right)^3\right)^5 = \left(1+\frac{1}{x}\right)^{15}$ Explanation: Recognizing this algebraic identity significantly simplifies the expression, making subsequent calculations feasible. It transforms a complex polynomial term into a simpler binomial power.

Step 2: Rewrite the entire expression.

Now substitute the simplified third term back into the original expression: $E = (1+x)(1-x^2)\left(1+\frac{1}{x}\right)^{15}$ Next, we apply the difference of squares identity to the second term, $(1-x^2) = (1-x)(1+x)$. We also rewrite $(1+\frac{1}{x})$ as $\frac{x+1}{x}$. $E = (1+x)(1-x)(1+x)\left(\frac{x+1}{x}\right)^{15}$ Combine the $(1+x)$ terms and distribute the exponent: $E = (1+x)^2 (1-x) \frac{(x+1)^{15}}{x^{15}}$ Combine the $(1+x)^2$ and $(x+1)^{15}$ terms (since $(x+1) = (1+x)$) to get $(1+x)^{17}$. $E = \frac{(1+x)^{17}(1-x)}{x^{15}}$ Distribute the $(1-x)$ term in the numerator: $E = \frac{(1+x)^{17} - x(1+x)^{17}}{x^{15}}$ Explanation: These steps are taken to combine all factors into a single, manageable form, which is essential for easily extracting coefficients using the Binomial Theorem. By expressing the entire numerator as a function of $(1+x)$ and the denominator as a power of $x$, we set up the problem for straightforward coefficient extraction.

Step 3: Find the coefficient of $x^3$.

We need the coefficient of $x^3$ in the expansion of $E = \frac{(1+x)^{17} - x(1+x)^{17}}{x^{15}}$. To find the coefficient of $x^3$ in $E$, we need to find the coefficient of $x^3 \cdot x^{15} = x^{18}$ in the numerator, which is $(1+x)^{17} - x(1+x)^{17}$.

Let's consider the two parts of the numerator:

1. Coefficient of $x^{18}$ in $(1+x)^{17}$: The general term in the expansion of $(1+x)^{17}$ is $\binom{17}{k}x^k$. We are looking for the coefficient of $x^{18}$, so $k=18$. However, the maximum power of $x$ in $(1+x)^{17}$ is $x^{17}$ (when $k=17$). Since $18 > 17$, the coefficient of $x^{18}$ in $(1+x)^{17}$ is 0. Explanation: The power of $x$ cannot exceed the exponent of the binomial $(1+x)^n$.

2. Coefficient of $x^{18}$ in $-x(1+x)^{17}$: This is equivalent to finding the coefficient of $x^{17}$ in $(1+x)^{17}$ and then multiplying by $-1$. The coefficient of $x^{17}$ in $(1+x)^{17}$ is $\binom{17}{17}$. $\binom{17}{17} = 1$. So, the coefficient from this part is $-1 \cdot 1 = -1$. Explanation: To find the coefficient of $x^m$ in $x^p f(x)$, we need the coefficient of $x^{m-p}$ in $f(x)$. Here, $m=18$, $p=1$, and $f(x)=(1+x)^{17}$.

Summing these two parts, the coefficient of $x^3$ in $E$ is $0 + (-1) = \mathbf{-1}$.

Step 4: Find the coefficient of $x^{-13}$.

We need the coefficient of $x^{-13}$ in $E = \frac{(1+x)^{17} - x(1+x)^{17}}{x^{15}}$. To find the coefficient of $x^{-13}$ in $E$, we need to find the coefficient of $x^{-13} \cdot x^{15} = x^2$ in the numerator, which is $(1+x)^{17} - x(1+x)^{17}$.

Let's consider the two parts of the numerator:

1. Coefficient of $x^2$ in $(1+x)^{17}$: Using the Binomial Theorem, the coefficient of $x^2$ in $(1+x)^{17}$ is $\binom{17}{2}$. $\binom{17}{2} = \frac{17 \times (17-1)}{2 \times 1} = \frac{17 \times 16}{2} = 17 \times 8 = 136$. Explanation: This is a direct application of the Binomial Theorem for $(1+x)^n$.

2. Coefficient of $x^2$ in $-x(1+x)^{17}$: This is equivalent to finding the coefficient of $x^1$ in $(1+x)^{17}$ and then multiplying by $-1$. The coefficient of $x^1$ in $(1+x)^{17}$ is $\binom{17}{1}$. $\binom{17}{1} = 17$. So, the coefficient from this part is $-1 \cdot 17 = -17$. Explanation: Similar to Step 3, we adjust the target power due to the multiplicative factor $x$.

Summing these two parts, the coefficient of $x^{-13}$ in $E$ is $136 + (-17) = \mathbf{119}$.

Step 5: Calculate the sum of the coefficients.

The problem asks for the sum of the coefficients of $x^3$ and $x^{-13}$. Sum = (Coefficient of $x^3$) + (Coefficient of $x^{-13}$) Sum = $-1 + 119 = \mathbf{118}$.

3. Tips and Common Mistakes to Avoid

• Algebraic Simplification First: Always look for opportunities to simplify the expression using algebraic identities before diving into binomial expansions. This can drastically reduce the complexity of the problem.
• Handling Negative Powers/Division: When an expression is of the form $\frac{P(x)}{x^k}$, and you need the coefficient of $x^m$, you effectively need the coefficient of $x^{m+k}$ in $P(x)$. Be very careful with the signs and arithmetic.
• Binomial Coefficient Properties: Remember that $\binom{n}{k} = 0$ if $k < 0$ or $k > n$. This is crucial for terms that might appear to be beyond the highest power in a binomial expansion.
• Careful with Signs: A single sign error, as seen in the subtraction of the coefficients, can lead to an incorrect final answer. Double-check all multiplications and subtractions.

4. Summary and Key Takeaway

This problem is an excellent example of how combining algebraic simplification with the Binomial Theorem can solve seemingly complex problems. The key takeaway is to methodically simplify the given expression first, especially by recognizing standard identities like $(a+b)^3$ and $(a-b)(a+b)$. Once the expression is in a form like $\frac{A(x)}{x^k}$, finding the coefficient of $x^m$ requires finding the coefficient of $x^{m+k}$ in the polynomial $A(x)$. This systematic approach ensures accuracy and clarity in solving such problems.