(x+3)^3 Binomial Theorem

Jasonbradley

3 min read

·

Jun 16, 2024

24

Share

Expanding (x+3)³ Using the Binomial Theorem

The binomial theorem provides a powerful way to expand expressions of the form (x + y)ⁿ for any positive integer n. Let's explore how to apply it to expand (x+3)³.

Understanding the Binomial Theorem

The binomial theorem states that:

(x + y)ⁿ = ∑_(k=0)^n (n choose k) x^(n-k) y^k

Where:

• (n choose k) represents the binomial coefficient, calculated as n! / (k! * (n-k)!). This represents the number of ways to choose k objects from a set of n objects.
• ∑_(k=0)^n represents the sum from k = 0 to n.

Applying the Binomial Theorem to (x+3)³

Let's break down the expansion of (x+3)³ using the binomial theorem:

1. Identify n: In this case, n = 3.

2. Calculate the binomial coefficients:

   • (3 choose 0) = 3! / (0! * 3!) = 1
   • (3 choose 1) = 3! / (1! * 2!) = 3
   • (3 choose 2) = 3! / (2! * 1!) = 3
   • (3 choose 3) = 3! / (3! * 0!) = 1

3. Apply the formula:

(x + 3)³ = ∑_(k=0)^3 (3 choose k) x^(3-k) 3^k

4. Expand the sum:

(x + 3)³ = (3 choose 0) x³ 3⁰ + (3 choose 1) x² 3¹ + (3 choose 2) x¹ 3² + (3 choose 3) x⁰ 3³

5. Simplify:

(x + 3)³ = 1 * x³ * 1 + 3 * x² * 3 + 3 * x * 9 + 1 * 1 * 27

(x + 3)³ = x³ + 9x² + 27x + 27

Conclusion

Therefore, the expanded form of (x + 3)³ using the binomial theorem is x³ + 9x² + 27x + 27. This demonstrates the power of the binomial theorem to systematically expand expressions raised to any positive integer power.