To expand $(x + 1)^3$, we can use the binomial theorem or direct multiplication. Here’s the step-by-step expansion:

1. Use the binomial expansion formula: $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ For $a = x$, $b = 1$, and $n = 3$: $(x + 1)^3 = \binom{3}{0} x^3 \cdot 1^0 + \binom{3}{1} x^2 \cdot 1^1 + \binom{3}{2} x^1 \cdot 1^2 + \binom{3}{3} x^0 \cdot 1^3$

2. Calculate the binomial coefficients: $\binom{3}{0} = 1, \quad \binom{3}{1} = 3, \quad \binom{3}{2} = 3, \quad \binom{3}{3} = 1$

3. Substituting back, we get: $(x + 1)^3 = 1 \cdot x^3 + 3 \cdot x^2 + 3 \cdot x + 1$ Thus, $(x + 1)^3 = x^3 + 3x^2 + 3x + 1$

Therefore, the expansion of $(x + 1)^3$ is: $\boxed{x^3 + 3x^2 + 3x + 1}$

If you would like further details or have any questions, let me know! Here are some related questions you might consider:

1. How can we derive the binomial theorem?
2. What is the significance of binomial coefficients in combinatorics?
3. Can you provide examples of using the binomial theorem with different values of $n$?
4. How does this expansion relate to polynomial functions?
5. What are the applications of polynomial expansions in calculus?

Tip: Practicing expansions with different values will strengthen your understanding of polynomial behavior!