Binomial expansion

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

In this case, $a = 2x$, $b = -3y$, and $n = 4$.

For $k=0$:

$$\binom{4}{0} (2x)^{4-0} (-3y)^0 = 1 \cdot (2x)^4 \cdot 1 = 16x^4$$

For $k=1$:

$$\binom{4}{1} (2x)^{4-1} (-3y)^1 = 4 \cdot (2x)^3 \cdot (-3y) = 4 \cdot 8x^3 \cdot (-3y) = -96x^3y$$

For $k=2$:

$$\binom{4}{2} (2x)^{4-2} (-3y)^2 = 6 \cdot (2x)^2 \cdot (9y^2) = 6 \cdot 4x^2 \cdot 9y^2 = 216x^2y^2$$

For $k=3$:

$$\binom{4}{3} (2x)^{4-3} (-3y)^3 = 4 \cdot (2x)^1 \cdot (-27y^3) = 4 \cdot 2x \cdot (-27y^3) = -216xy^3$$

For $k=4$:

$$\binom{4}{4} (2x)^{4-4} (-3y)^4 = 1 \cdot (2x)^0 \cdot (81y^4) = 1 \cdot 1 \cdot 81y^4 = 81y^4$$

$$(2x - 3y)^4 = 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4$$

The final answer is $16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4$.