## The Proof of (a + b)³

The expansion of (a + b)³ is a fundamental concept in algebra. It is often used in various mathematical contexts and can be derived using two main methods:

**1. Using the Distributive Property**

This method involves repeatedly applying the distributive property of multiplication over addition:

**Step 1:** Expand the cube of the binomial.
(a + b)³ = (a + b) * (a + b) * (a + b)

**Step 2:** Expand the first two factors using the distributive property.
(a + b) * (a + b) = a(a + b) + b(a + b) = a² + ab + ba + b² = a² + 2ab + b²

**Step 3:** Multiply the result from Step 2 by (a + b).
(a² + 2ab + b²) * (a + b) = a(a² + 2ab + b²) + b(a² + 2ab + b²)

**Step 4:** Apply the distributive property again.
a(a² + 2ab + b²) + b(a² + 2ab + b²) = a³ + 2a²b + ab² + ba² + 2ab² + b³

**Step 5:** Combine like terms.
a³ + 2a²b + ab² + ba² + 2ab² + b³ = **a³ + 3a²b + 3ab² + b³**

Therefore, (a + b)³ = **a³ + 3a²b + 3ab² + b³**

**2. Using the Binomial Theorem**

The binomial theorem provides a general formula for expanding any binomial raised to a power. It states:

**(a + b)ⁿ = ∑(k=0 to n) (n choose k) * a^(n-k) * b^k**

where (n choose k) represents the binomial coefficient, calculated as n!/(k!(n-k)!).

**Step 1:** Apply the binomial theorem to (a + b)³.

(a + b)³ = ∑(k=0 to 3) (3 choose k) * a^(3-k) * b^k

**Step 2:** Expand the summation.

(a + b)³ = (3 choose 0) * a³ * b⁰ + (3 choose 1) * a² * b¹ + (3 choose 2) * a¹ * b² + (3 choose 3) * a⁰ * b³

**Step 3:** Calculate the binomial coefficients.

(3 choose 0) = 1, (3 choose 1) = 3, (3 choose 2) = 3, (3 choose 3) = 1

**Step 4:** Substitute the coefficients and simplify.

(a + b)³ = 1 * a³ * 1 + 3 * a² * b + 3 * a * b² + 1 * 1 * b³ = **a³ + 3a²b + 3ab² + b³**

Therefore, using both methods, we have shown that:

**(a + b)³ = a³ + 3a²b + 3ab² + b³**

This expansion is useful for various algebraic manipulations, and it forms the basis for further expansions of higher powers of binomials.