### Step-by-Step Solution
#### 3. Find the expansion of [tex]\((x + y)^4\)[/tex].
To expand [tex]\((x + y)^4\)[/tex], we can use the binomial theorem. The binomial theorem states:
[tex]\[
(a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k
\][/tex]
In our case, [tex]\(a = x\)[/tex], [tex]\(b = y\)[/tex], and [tex]\(n = 4\)[/tex]. Applying the binomial theorem, we get:
[tex]\[
(x + y)^4 = \sum_{k=0}^4 \binom{4}{k} x^{4-k} y^k
\][/tex]
Now we can expand it term by term:
- When [tex]\(k = 0\)[/tex]:
[tex]\[
\binom{4}{0} x^{4-0} y^0 = \binom{4}{0} x^4 = 1 \cdot x^4 = x^4
\][/tex]
- When [tex]\(k = 1\)[/tex]:
[tex]\[
\binom{4}{1} x^{4-1} y^1 = \binom{4}{1} x^3 y = 4 \cdot x^3 y = 4x^3 y
\][/tex]
- When [tex]\(k = 2\)[/tex]:
[tex]\[
\binom{4}{2} x^{4-2} y^2 = \binom{4}{2} x^2 y^2 = 6 \cdot x^2 y^2 = 6x^2 y^2
\][/tex]
- When [tex]\(k = 3\)[/tex]:
[tex]\[
\binom{4}{3} x^{4-3} y^3 = \binom{4}{3} x y^3 = 4 \cdot x y^3 = 4x y^3
\][/tex]
- When [tex]\(k = 4\)[/tex]:
[tex]\[
\binom{4}{4} x^{4-4} y^4 = \binom{4}{4} y^4 = 1 \cdot y^4 = y^4
\][/tex]
Putting all the terms together, the full expansion is:
[tex]\[
(x + y)^4 = x^4 + 4x^3 y + 6x^2 y^2 + 4x y^3 + y^4
\][/tex]
#### 4. Find the coefficient of [tex]\(x^3\)[/tex] in the expansion of [tex]\((x + y)^4\)[/tex].
From the expansion we found:
[tex]\[
(x + y)^4 = x^4 + 4x^3 y + 6x^2 y^2 + 4x y^3 + y^4
\][/tex]
We need to identify the coefficient of the term that includes [tex]\(x^3\)[/tex]. Looking at the term with [tex]\(x^3\)[/tex]:
- The term [tex]\(4x^3 y\)[/tex] contains [tex]\(x^3\)[/tex].
The coefficient of [tex]\(x^3\)[/tex] in [tex]\(4x^3 y\)[/tex] is:
[tex]\[
4y
\][/tex]
If [tex]\(y\)[/tex] is not specified, the coefficient is symbolic and given as [tex]\(4y\)[/tex]. If we are considering this strictly as a numeric value:
[tex]\[
\coeff_x3 = 0
\][/tex]
Thus, the coefficient of [tex]\(x^3\)[/tex] is [tex]\(4\)[/tex].
