The given expression is $(2x + 3y)^6$, and we are asked to expand it using the binomial theorem.

The binomial theorem states that:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

In this case:

• $a = 2x$
• $b = 3y$
• $n = 6$

Thus, we apply the binomial expansion formula:

$(2x + 3y)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^{6-k} (3y)^k$

We will compute each term:

1. For $k = 0$:
   $\binom{6}{0} (2x)^6 (3y)^0 = 1 \cdot (64x^6) \cdot 1 = 64x^6$

2. For $k = 1$:
   $\binom{6}{1} (2x)^5 (3y)^1 = 6 \cdot (32x^5) \cdot (3y) = 576x^5y$

3. For $k = 2$:
   $\binom{6}{2} (2x)^4 (3y)^2 = 15 \cdot (16x^4) \cdot (9y^2) = 2160x^4y^2$

4. For $k = 3$:
   $\binom{6}{3} (2x)^3 (3y)^3 = 20 \cdot (8x^3) \cdot (27y^3) = 4320x^3y^3$

5. For $k = 4$:
   $\binom{6}{4} (2x)^2 (3y)^4 = 15 \cdot (4x^2) \cdot (81y^4) = 4860x^2y^4$

6. For $k = 5$:
   $\binom{6}{5} (2x)^1 (3y)^5 = 6 \cdot (2x) \cdot (243y^5) = 2916xy^5$

7. For $k = 6$:
   $\binom{6}{6} (2x)^0 (3y)^6 = 1 \cdot (1) \cdot (729y^6) = 729y^6$

Now we combine all these terms to obtain the full expansion:

$(2x + 3y)^6 = 64x^6 + 576x^5y + 2160x^4y^2 + 4320x^3y^3 + 4860x^2y^4 + 2916xy^5 + 729y^6$

This is the expanded and simplified form of the expression.

Would you like more details on any part of this expansion? Let me know if you have questions!

Here are 5 related questions for practice:

1. How would you expand $(x + 2y)^5$ using the binomial theorem?
2. What is the general formula for the binomial coefficient $\binom{n}{k}$?
3. How would the expansion differ if the exponent in the original problem were 7 instead of 6?
4. Can you apply the binomial theorem to expand $(a - b)^6$?
5. How can the binomial theorem be applied to solve real-world problems in probability or combinatorics?

Tip: The binomial coefficients $\binom{n}{k}$ can also be found using Pascal's Triangle!