To find the coefficient of $x^7$ in the expansion of $(1 - x - x^2 + x^3)^6$, we can use the multinomial theorem. We need to expand $(1 - x - x^2 + x^3)^6$ and identify the terms that contribute to $x^7$.

Step 1: Express the expansion in terms of the multinomial theorem.

The multinomial theorem for expanding $(a + b + c + d)^n$ gives the general term:

$\text{Term} = \frac{6!}{k_1!k_2!k_3!k_4!} \cdot a^{k_1} b^{k_2} c^{k_3} d^{k_4}$ where $k_1 + k_2 + k_3 + k_4 = 6$, and the exponents of each term are powers of $1$, $-x$, $-x^2$, and $x^3$.

In our case:

• $a = 1$,
• $b = -x$,
• $c = -x^2$,
• $d = x^3$.

Step 2: Find combinations that contribute to $x^7$.

Each of the terms $1$, $-x$, $-x^2$, and $x^3$ contributes to different powers of $x$. We need to find the combinations of powers $k_1$, $k_2$, $k_3$, and $k_4$ such that:

$k_2 + 2k_3 + 3k_4 = 7 \quad \text{and} \quad k_1 + k_2 + k_3 + k_4 = 6.$

We'll now determine all possible sets of non-negative integers $(k_1, k_2, k_3, k_4)$ that satisfy these conditions.

Step 3: Solve the system of equations.

Let's find the possible values of $k_2$, $k_3$, and $k_4$ that satisfy the equation $k_2 + 2k_3 + 3k_4 = 7$ under the constraint that $k_1 + k_2 + k_3 + k_4 = 6$.

Possible combinations:

• $k_2 = 1$, $k_3 = 2$, $k_4 = 1$
• $k_2 = 4$, $k_3 = 1$, $k_4 = 1$
• $k_2 = 7$, ( k_