Suppose $z^{3}=2+2i$, where $i=\sqrt{-1}$. The product of all possible values of the real part of $z$ can be written in the form $\frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers. Find $p+q$.
1. **Identify the roots using De Moivre's Theorem:**
   Given \( z^3 = 2 + 2i \), we first convert \( 2 + 2i \) to polar form. The magnitude \( r \) and argument \( \theta \) are calculated as follows:
   \[
   r = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}
   \]
   \[
   \theta = \tan^{-1}\left(\frac{2}{2}\right) = \tan^{-1}(1) = \frac{\pi}{4}
   \]
   Thus, \( 2 + 2i = 2\sqrt{2} \text{cis} \left( \frac{\pi}{4} \right) \).

2. **Apply De Moivre's Theorem to find the roots:**
   We need to find the cube roots of \( 2\sqrt{2} \text{cis} \left( \frac{\pi}{4} \right) \). Using De Moivre's Theorem, the roots are given by:
   \[
   z_k = \sqrt[3]{2\sqrt{2}} \text{cis} \left( \frac{\frac{\pi}{4} + 2k\pi}{3} \right) \quad \text{for} \quad k = 0, 1, 2
   \]
   Simplifying \( \sqrt[3]{2\sqrt{2}} \):
   \[
   \sqrt[3]{2\sqrt{2}} = \sqrt[3]{2 \cdot 2^{1/2}} = \sqrt[3]{2^{3/2}} = 2^{1/2} = \sqrt{2}
   \]
   Therefore, the roots are:
   \[
   z_k = \sqrt{2} \text{cis} \left( \frac{\pi}{12} + \frac{2k\pi}{3} \right) \quad \text{for} \quad k = 0, 1, 2
   \]

3. **Calculate the real parts of the roots:**
   The real parts of the roots are given by:
   \[
   \text{Re}(z_k) = \sqrt{2} \cos \left( \frac{\pi}{12} + \frac{2k\pi}{3} \right)
   \]
   For \( k = 0 \):
   \[
   \text{Re}(z_0) = \sqrt{2} \cos \left( \frac{\pi}{12} \right)
   \]
   For \( k = 1 \):
   \[
   \text{Re}(z_1) = \sqrt{2} \cos \left( \frac{\pi}{12} + \frac{2\pi}{3} \right) = \sqrt{2} \cos \left( \frac{9\pi}{12} \right) = \sqrt{2} \cos \left( \frac{3\pi}{4} \right)
   \]
   For \( k = 2 \):
   \[
   \text{Re}(z_2) = \sqrt{2} \cos \left( \frac{\pi}{12} + \frac{4\pi}{3} \right) = \sqrt{2} \cos \left( \frac{17\pi}{12} \right)
   \]

4. **Use trigonometric identities to find the product of the real parts:**
   Using the known values:
   \[
   \cos \left( \frac{\pi}{12} \right) = \frac{\sqrt{6} + \sqrt{2}}{4}
   \]
   \[
   \cos \left( \frac{3\pi}{4} \right) = -\frac{\sqrt{2}}{2}
   \]
   \[
   \cos \left( \frac{17\pi}{12} \right) = -\cos \left( \frac{5\pi}{12} \right) = -\left( \frac{\sqrt{6} - \sqrt{2}}{4} \right)
   \]
   The product of the real parts is:
   \[
   \left( \sqrt{2} \cos \left( \frac{\pi}{12} \right) \right) \left( \sqrt{2} \cos \left( \frac{3\pi}{4} \right) \right) \left( \sqrt{2} \cos \left( \frac{17\pi}{12} \right) \right)
   \]
   \[
   = \left( \sqrt{2} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} \right) \left( \sqrt{2} \cdot -\frac{\sqrt{2}}{2} \right) \left( \sqrt{2} \cdot -\frac{\sqrt{6} - \sqrt{2}}{4} \right)
   \]
   \[
   = \left( \frac{\sqrt{12} + \sqrt{4}}{4} \right) \left( -\frac{2}{2} \right) \left( -\frac{\sqrt{12} - \sqrt{4}}{4} \right)
   \]
   \[
   = \left( \frac{2\sqrt{3} + 2}{4} \right) \left( -1 \right) \left( -\frac{2\sqrt{3} - 2}{4} \right)
   \]
   \[
   = \left( \frac{\sqrt{3} + 1}{2} \right) \left( \frac{\sqrt{3} - 1}{2} \right)
   \]
   \[
   = \frac{(\sqrt{3} + 1)(\sqrt{3} - 1)}{4} = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2}
   \]

5. **Express the product in the form \(\frac{p}{q}\) and find \(p + q\):**
   The product of all possible values of the real part of \( z \) is \( \frac{1}{2} \). Here, \( p = 1 \) and \( q = 2 \), so \( p + q = 1 + 2 = 3 \).

The final answer is \( \boxed{3} \)