Question:

**Step 1: Simplify the base expression** The expression inside the brackets is $1 - x - x^2(1-x)$. $$1 - x - x^2(1-x) = 1 - x - x^2 + x^3$$ The problem requires finding the coefficient of $x^7$ in the expansion of $[1 - x - x^2 + x^3]^6$. **Step 2: Apply the multinomial theorem** The multinomial expansion of $(a+b+c+d)^N$ is given by $$\sum_{n_0+n_1+n_2+n_3=N} \frac{N!}{n_0!n_1!n_2!n_3!} a^{n_0} b^{n_1} c^{n_2} d^{n_3}$$ In this case, $N=6$, and the terms are $1$, $-x$, $-x^2$, and $x^3$. The general term in the expansion of $[1 + (-x) + (-x^2) + (x^3)]^6$ is: $$\frac{6!}{n_0!n_1!n_2!n_3!} (1)^{n_0} (-x)^{n_1} (-x^2)^{n_2} (x^3)^{n_3}$$ $$= \frac{6!}{n_0!n_1!n_2!n_3!} (1)^{n_0} (-1)^{n_1} x^{n_1} (-1)^{n_2} x^{2n_2} (1)^{n_3} x^{3n_3}$$ $$= \frac{6!}{n_0!n_1!n_2!n_3!} (-1)^{n_1+n_2} x^{n_1+2n_2+3n_3}$$ **Step 3: Determine conditions for the coefficients** We seek the coefficient of $x^7$, so the power of $x$ must be 7: $$n_1 + 2n_2 + 3n_3 = 7$$ The sum of the exponents of the terms must be 6: $$n_0 + n_1 + n_2 + n_3 = 6$$ All $n_i$ must be non-negative integers ($n_i \ge 0$). **Step 4: Find valid combinations of $(n_0, n_1, n_2, n_3)$ and calculate their coefficients** We systematically find all combinations of $(n_0, n_1, n_2, n_3)$ that satisfy both conditions. * **Case A: $n_3 = 0$** The conditions become: $n_1 + 2n_2 = 7$ $n_0 + n_1 + n_2 = 6 \implies n_0 = 6 - n_1 - n_2$ * If $n_2 = 0$, $n_1 = 7$. Then $n_0 = 6 - 7 - 0 = -1$, which is not allowed. * If $n_2 = 1$, $n_1 = 5$. Then $n_0 = 6 - 5 - 1 = 0$. Combination: $(0, 5, 1, 0)$. Coefficient: $\frac{6!}{0!5!1!0!} (-1)^{5+1} = \frac{720}{1 \cdot 120 \cdot 1 \cdot 1} (1) = 6$. * If $n_2 = 2$, $n_1 = 3$. Then $n_0 = 6 - 3 - 2 = 1$. Combination: $(1, 3, 2, 0)$. Coefficient: $\frac{6!}{1!3!2!0!} (-1)^{3+2} = \frac{720}{1 \cdot 6 \cdot 2 \cdot 1} (-1) = -60$. * If $n_2 = 3$, $n_1 = 1$. Then $n_0 = 6 - 1 - 3 = 2$. Combination: $(2, 1, 3, 0)$. Coefficient: $\frac{6!}{2!1!3!0!} (-1)^{1+3} = \frac{720}{2 \cdot 1 \cdot 6 \cdot 1} (1) = 60$. * If $n_2 \ge 4$, $n_1$ would be negative. * **Case B: $n_3 = 1$** The conditions become: $n_1 + 2n_2 + 3(1) = 7 \implies n_1 + 2n_2 = 4$ $n_0 + n_1 + n_2 + 1 = 6 \implies n_0 = 5 - n_1 - n_2$ * If $n_2 = 0$, $n_1 = 4$. Then $n_0 = 5 - 4 - 0 = 1$. Combination: $(1, 4, 0, 1)$. Coefficient: $\frac{6!}{1!4!0!1!} (-1)^{4+0} = \frac{720}{1 \cdot 24 \cdot 1 \cdot 1} (1) = 30$. * If $n_2 = 1$, $n_1 = 2$. Then $n_0 = 5 - 2 - 1 = 2$. Combination: $(2, 2, 1, 1)$. Coefficient: $\frac{6!}{2!2!1!1!} (-1)^{2+1} = \frac{720}{2 \cdot 2 \cdot 1 \cdot 1} (-1) = -180$. * If $n_2 = 2$, $n_1 = 0$. Then $n_0 = 5 - 0 - 2 = 3$. Combination: $(3, 0, 2, 1)$. Coefficient: $\frac{6!}{3!0!2!1!} (-1)^{0+2} = \frac{720}{6 \cdot 1 \cdot 2 \cdot 1} (1) = 60$. * If $n_2 \ge 3$, $n_1$ would be negative. * **Case C: $n_3 = 2$** The conditions become: $n_1 + 2n_2 + 3(2) = 7 \implies n_1 + 2n_2 = 1$ $n_0 + n_1 + n_2 + 2 = 6 \implies n_0 = 4 - n_1 - n_2$ * If $n_2 = 0$, $n_1 = 1$. Then $n_0 = 4 - 1 - 0 = 3$. Combination: $(3, 1, 0, 2)$. Coefficient: $\frac{6!}{3!1!0!2!} (-1)^{1+0} = \frac{720}{6 \cdot 1 \cdot 1 \cdot 2} (-1) = -60$. * If $n_2 \ge 1$, $n_1$ would be negative. * **Case D: $n_3 \ge 3$** If $n_3 = 3$, then $3n_3 = 9$, which is greater than 7. Thus, no more combinations are possible. **Step 5: Sum all contributions** The total coefficient of $x^7$ is the sum of the coefficients from all valid combinations: $$6 + (-60) + 60 + 30 + (-180) + 60 + (-60)$$ $$= 6 - 60 + 60 + 30 - 180 + 60 - 60$$ $$= 6 + 30 - 180$$ $$= 36 - 180$$ $$= -144$$