1. The problem is to expand the binomial expression $(y-3)^4$. 2. Recall the binomial theorem for expansion: $$ (a - b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} (-b)^k $$ 3. Here, $a = y$, $b = 3$, and $n = 4$. 4. Calculate each term using the binomial coefficients ${4 \choose k}$: - When $k=0$: ${4 \choose 0} y^4 (-3)^0 = 1 \cdot y^4 \cdot 1 = y^4$ - When $k=1$: ${4 \choose 1} y^3 (-3)^1 = 4 \cdot y^3 \cdot (-3) = -12 y^3$ - When $k=2$: ${4 \choose 2} y^2 (-3)^2 = 6 \cdot y^2 \cdot 9 = 54 y^2$ - When $k=3$: ${4 \choose 3} y^1 (-3)^3 = 4 \cdot y \cdot (-27) = -108 y$ - When $k=4$: ${4 \choose 4} y^0 (-3)^4 = 1 \cdot 1 \cdot 81 = 81$ 5. Now, combine all terms: $$ (y-3)^4 = y^4 -12 y^3 + 54 y^2 - 108 y + 81 $$ This is the fully expanded expression.