To find the coefficient of $x^3$ in the expansion of $(1 - x + x^2)^5$, we can use the multinomial theorem.

Step 1: Understand the expression

The expression is $(1 - x + x^2)^5$. We are interested in finding the coefficient of $x^3$ when it is expanded.

Step 2: Apply the multinomial expansion

The general form of the multinomial expansion for $(a + b + c)^n$ is:

$(a + b + c)^n = \sum_{i+j+k=n} \frac{n!}{i!j!k!} a^i b^j c^k$ where $i + j + k = n$, and $i, j, k \geq 0$.

In our case, $a = 1$, $b = -x$, and $c = x^2$, and we need to find the coefficient of $x^3$ in the expansion of $(1 - x + x^2)^5$.

Step 3: Identify the relevant terms

We need to find the terms where the powers of $x$ sum to 3. For any term in the expansion, the exponent of $x$ is $j + 2k$ because:

• The $b = -x$ contributes $j$ factors of $x$.
• The $c = x^2$ contributes $2k$ factors of $x$.

Thus, we need:

$j + 2k = 3$

Step 4: Solve for the possible values of $j$ and $k$

We now solve for all possible combinations of $j$ and $k$ such that $j + 2k = 3$:

1. If $k = 1$, then $j + 2(1) = 3$, so $j = 1$.
2. If $k = 0$, then $j + 2(0) = 3$, so $j = 3$.

Thus, the possible combinations are:

• $j = 1$, $k = 1$
• $j = 3$, $k = 0$

Step 5: Calculate the coefficients

For each combination, the corresponding multinomial term is given by:

$\frac{5!}{i!j!k!} \cdot 1^i \cdot (-x)^j \cdot (x^2)^k$ where $i + j + k = 5$.

1. For $j = 1$ and $k = 1$:

   • $i = 3$ because $i + j + k = 5$.
   • The term is: $\frac{5!}{3!1!1!} \cdot (-x)^1 \cdot (x^2)^1 = \frac{120}{6 \cdot 1 \cdot 1} \cdot (-x) \cdot x^2 = 20 \cdot (-x) \cdot x^2 = -20x^3$
   • The contribution to $x^3$ is $-20$.

2. For $j = 3$ and $k = 0$:

   • $i = 2$ because $i + j + k = 5$.
   • The term is: $\frac{5!}{2!3!0!} \cdot (-x)^3 \cdot (x^2)^0 = \frac{120}{2 \cdot 6 \cdot 1} \cdot (-x)^3 = 10 \cdot (-x^3) = -10x^3$
   • The contribution to $x^3$ is $-10$.

Step 6: Combine the contributions

The total coefficient of $x^3$ is the sum of the contributions from both cases:

$-20 + (-10) = -30$

Final Answer:

The coefficient of $x^3$ in the expansion of $(1 - x + x^2)^5$ is $\boxed{-30}$.

Would you like further explanation or assistance with another problem?

Additional Questions:

1. How can we apply the multinomial theorem to expansions of other powers like $(1 + x + x^2)^4$?
2. What is the general form for finding the coefficient of $x^n$ in a multinomial expansion?
3. Can you expand $(1 - 2x + x^3)^5$ and find the coefficient of $x^3$?
4. What is the difference between the binomial and multinomial expansions?
5. How does the symmetry of the multinomial coefficients affect the expansion of polynomials?

Tip:

When using the multinomial expansion, always pay close attention to the powers of each term (whether linear, quadratic, etc.) and ensure you account for all possible combinations of exponents that match the desired total.