Comprehensive Solution

1. Key Concept: The Binomial Theorem

The Binomial Theorem provides a formula for expanding algebraic expressions of the form $(a+b)^n$ into a sum of terms. The general form is: $(a+b)^n = \sum_{k=0}^{n} {^nC_k} a^{n-k} b^k$ where ${^nC_k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient, representing the number of ways to choose $k$ items from a set of $n$ items.

For a term involving $(1-x)^n$, the expansion becomes $(1-x)^n = \sum_{k=0}^{n} {^nC_k} (1)^{n-k} (-x)^k = \sum_{k=0}^{n} {^nC_k} (-1)^k x^k$. When dealing with a product of two binomial expansions, say $(A(x) \times B(x))$, to find the coefficient of a specific power of $x$, we identify pairs of terms from each expansion whose powers of $x$ sum up to the desired total power.

2. Simplification of the Given Expression

The first crucial step is to simplify the given expression $(1 - x - x^2 + x^3)^6$. This is a multinomial, which is generally harder to expand directly. We look for common factors: $1 - x - x^2 + x^3$ We can group terms: $= (1 - x) - (x^2 - x^3)$ $= (1 - x) - x^2(1 - x)$ Now, we can factor out the common term $(1 - x)$: $= (1 - x)(1 - x^2)$

Therefore, the original expression can be rewritten as: ${\left( {1 - x - {x^2} + {x^3}} \right)^6} = {\left[ {\left( {1 - x} \right)\left( {1 - {x^2}} \right)} \right]^6} = {\left( {1 - x} \right)^6}{\left( {1 - {x^2}} \right)^6}$

Why this step is taken: This factorization transforms a complex multinomial expression into a product of two simpler binomial expressions. It's much easier to work with $(1-x)^6$ and $(1-x^2)^6$ separately and then combine their terms.

3. Expanding Each Binomial Term (General Term)

Now we need to find the coefficient of ${x^7}$ in the product of these two expansions. Let's write the general term for each:

• For the expansion of $(1 - x)^6$: The general term, $T_{r+1}$, is given by ${^6C_r} (1)^{6-r} (-x)^r = {^6C_r} (-1)^r x^r$. Let $C_r^{(1)}$ denote the coefficient of $x^r$ in $(1-x)^6$, so $C_r^{(1)} = {^6C_r} (-1)^r$.

• For the expansion of $(1 - x^2)^6$: The general term, $T_{s+1}$, is given by ${^6C_s} (1)^{6-s} (-x^2)^s = {^6C_s} (-1)^s (x^2)^s = {^6C_s} (-1)^s x^{2s}$. Let $C_s^{(2)}$ denote the coefficient of $x^{2s}$ in $(1-x^2)^6$, so $C_s^{(2)} = {^6C_s} (-1)^s$.

Why this step is taken: Writing the general term helps us systematically identify coefficients for specific powers of $x$ and also correctly handle the alternating signs due to the $(1-y)^n$ form.

4. Identifying Relevant Combinations of Powers

We need the coefficient of $x^7$ in the product $(1-x)^6 (1-x^2)^6$. This means we need to find pairs of terms, one from each expansion, whose powers of $x$ multiply to $x^7$. If we take a term $x^r$ from $(1-x)^6$ and a term $x^{2s}$ from $(1-x^2)^6$, their product will have power $x^{r+2s}$. We need this sum to be 7: $r + 2s = 7$ Also, remember the constraints on $r$ and $s$: $0 \le r \le 6$ and $0 \le s \le 6$.

Let's list the possible pairs $(r, s)$ satisfying $r+2s=7$:

• If $s=0$: $r = 7 - 2(0) = 7$. This is not possible because $r$ must be $\le 6$.
• If $s=1$: $r = 7 - 2(1) = 5$. This is a valid pair: $(r, s) = (5, 1)$. (This means we combine $x^5$ from $(1-x)^6$ with $x^2$ from $(1-x^2)^6$)
• If $s=2$: $r = 7 - 2(2) = 3$. This is a valid pair: $(r, s) = (3, 2)$. (This means we combine $x^3$ from $(1-x)^6$ with $x^4$ from $(1-x^2)^6$)
• If $s=3$: $r = 7 - 2(3) = 1$. This is a valid pair: $(r, s) = (1, 3)$. (This means we combine $x^1$ from $(1-x)^6$ with $x^6$ from $(1-x^2)^6$)
• If $s=4$: $r = 7 - 2(4) = -1$. This is not possible because $r$ must be $\ge 0$.

So, there are three pairs of $(r,s)$ that contribute to the coefficient of $x^7$.

Why this step is taken: This systematic approach ensures that we consider all possible ways to form the desired power $x^7$ and don't miss any contributing terms.

5. Calculating Coefficients for Each Combination

Now we calculate the product of the coefficients $C_r^{(1)} \times C_s^{(2)}$ for each valid pair $(r, s)$.

Recall the binomial coefficients: ${^6C_0} = 1$ ${^6C_1} = 6$ ${^6C_2} = 15$ ${^6C_3} = 20$ ${^6C_4} = 15$ ${^6C_5} = 6$ ${^6C_6} = 1$

• Case 1: $(r, s) = (5, 1)$

  • Coefficient of $x^5$ in $(1-x)^6$: $C_5^{(1)} = {^6C_5} (-1)^5 = 6 \times (-1) = -6$.
  • Coefficient of $x^2$ (from $x^{2s}$) in $(1-x^2)^6$: $C_1^{(2)} = {^6C_1} (-1)^1 = 6 \times (-1) = -6$.
  • Contribution to $x^7$: $(-6) \times (-6) = 36$

• Case 2: $(r, s) = (3, 2)$

  • Coefficient of $x^3$ in $(1-x)^6$: $C_3^{(1)} = {^6C_3} (-1)^3 = 20 \times (-1) = -20$.
  • Coefficient of $x^4$ (from $x^{2s}$) in $(1-x^2)^6$: $C_2^{(2)} = {^6C_2} (-1)^2 = 15 \times (1) = 15$.
  • Contribution to $x^7$: $(-20) \times (15) = -300$

• Case 3: $(r, s) = (1, 3)$

  • Coefficient of $x^1$ in $(1-x)^6$: $C_1^{(1)} = {^6C_1} (-1)^1 = 6 \times (-1) = -6$.
  • Coefficient of $x^6$ (from $x^{2s}$) in $(1-x^2)^6$: $C_3^{(2)} = {^6C_3} (-1)^3 = 20 \times (-1) = -20$.
  • Contribution to $x^7$: $(-6) \times (-20) = 120$

Why this step is taken: Each product of coefficients represents the total coefficient of $x^7$ for a specific combination of terms from the two expansions. We must calculate these accurately, paying close attention to the signs.

6. Summing the Coefficients

To find the total coefficient of $x^7$, we sum the contributions from all valid cases: Total coefficient of $x^7 = 36 + (-300) + 120$ $= 36 - 300 + 120$ $= 156 - 300$ $= -144$

Why this step is taken: The coefficient of a specific term in an algebraic expansion is the sum of all ways that term can be formed through multiplication.

7. Tips for Success & Common Mistakes

• Factorization First: Always look for opportunities to factorize complex polynomial expressions into simpler binomial or polynomial forms before attempting expansion. This dramatically simplifies the problem.
• Sign Errors: Be extremely careful with signs, especially when dealing with expressions like $(1-x)^n$ or $(1-x^2)^n$. The ${(-1)^k}$ factor from the binomial expansion is crucial and frequently overlooked.
• Systematic Approach: When combining terms from multiple expansions, create a table or systematically list all possible combinations of powers that sum to the desired total. This prevents missing terms.
• Range of Indices: Always remember the limits on the indices ($r, s, k$) in binomial expansions ($0 \le k \le n$). This helps in filtering out invalid combinations.
• Binomial Coefficient Calculation: Double-check the calculation of binomial coefficients ${^nC_k}$. Common values for small $n$ can be memorized or quickly derived.

8. Summary and Key Takeaway

This problem effectively demonstrates how combining algebraic factorization with a systematic application of the Binomial Theorem can simplify the process of finding coefficients in complex expansions. The key is to break down the problem into manageable steps: simplify, identify general terms, find all contributing combinations, and then sum their coefficients. Accuracy in sign handling and binomial coefficient calculations is paramount.

The final answer is $-144$.