To express $(2x - 3y)^4$, we use the binomial theorem, which states that $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$. Here, $a = 2x$, $b = -3y$, and $n = 4$. Step 1: Write out the terms using the binomial theorem. The expansion will have $n+1 = 5$ terms: $$ \binom{4}{0}(2x)^4(-3y)^0 + \binom{4}{1}(2x)^3(-3y)^1 + \binom{4}{2}(2x)^2(-3y)^2 + \binom{4}{3}(2x)^1(-3y)^3 + \binom{4}{4}(2x)^0(-3y)^4 $$ Step 2: Calculate the binomial coefficients $\binom{n}{k}$. $$ \binom{4}{0} = 1 $$ $$ \binom{4}{1} = 4 $$ $$ \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 $$ $$ \binom{4}{3} = 4 $$ $$ \binom{4}{4} = 1 $$ Step 3: Expand each term. For $k=0$: $$ 1 \cdot (2x)^4 \cdot (-3y)^0 = 1 \cdot (16x^4) \cdot 1 = 16x^4 $$ For $k=1$: $$ 4 \cdot (2x)^3 \cdot (-3y)^1 = 4 \cdot (8x^3) \cdot (-3y) = -96x^3y $$ For $k=2$: $$ 6 \cdot (2x)^2 \cdot (-3y)^2 = 6 \cdot (4x^2) \cdot (9y^2) = 216x^2y^2 $$ For $k=3$: $$ 4 \cdot (2x)^1 \cdot (-3y)^3 = 4 \cdot (2x) \cdot (-27y^3) = -216xy^3 $$ For $k=4$: $$ 1 \cdot (2x)^0 \cdot (-3y)^4 = 1 \cdot 1 \cdot (81y^4) = 81y^4 $$ Step 4: Combine all the terms. $$ (2x - 3y)^4 = 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4 $$ The expanded expression is: $$ \boxed{16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4} $$