F) \( \left(\frac{y}{2}+3\right)^{3} \)

To expand \( \left(\frac{y}{2}+3\right)^{3} \), we can use the binomial theorem, which tells us that \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} \). Here, \( a = \frac{y}{2} \), \( b = 3 \), and \( n = 3 \). This gives us: \[ = \binom{3}{0} \left(\frac{y}{2}\right)^3 3^0 + \binom{3}{1} \left(\frac{y}{2}\right)^2 3^1 + \binom{3}{2} \left(\frac{y}{2}\right)^1 3^2 + \binom{3}{3} \left(\frac{y}{2}\right)^0 3^3 \] Calculating each term: 1. \( \binom{3}{0} \left(\frac{y}{2}\right)^3 3^0 = 1 \cdot \frac{y^3}{8} \cdot 1 = \frac{y^3}{8} \) 2. \( \binom{3}{1} \left(\frac{y}{2}\right)^2 3^1 = 3 \cdot \frac{y^2}{4} \cdot 3 = \frac{9y^2}{4} \) 3. \( \binom{3}{2} \left(\frac{y}{2}\right)^1 3^2 = 3 \cdot \frac{y}{2} \cdot 9 = \frac{27y}{2} \) 4. \( \binom{3}{3} \left(\frac{y}{2}\right)^0 3^3 = 1 \cdot 1 \cdot 27 = 27 \) Putting it all together: \[ \left(\frac{y}{2}+3\right)^{3} = \frac{y^3}{8} + \frac{9y^2}{4} + \frac{27y}{2} + 27 \] To write everything with a common denominator (which is 8): \[ = \frac{y^3}{8} + \frac{18y^2}{8} + \frac{108y}{8} + \frac{216}{8} \] So, the expanded form is: \[ = \frac{y^3 + 18y^2 + 108y + 216}{8} \]