Let $x$ be in the interval $\left(0, \frac{\pi}{2}\right)$ such that $\sin x - \cos x = \frac12$ . Then $\sin^3 x + \cos^3 x = \frac{m\sqrt{p}}{n}$ , where $m, n$, and $p$ are relatively prime positive integers, and $p$ is not divisible by the square of any prime. Find $m + n + p$.
1. Let $\sin x = a$ and $\cos x = b$. Given that $a - b = \frac{1}{2}$ and $a^2 + b^2 = 1$.
2. We need to find $\sin^3 x + \cos^3 x = a^3 + b^3$.
3. Recall the identity for the sum of cubes: 
   \[
   a^3 + b^3 = (a + b)(a^2 - ab + b^2)
   \]
4. First, solve for $a$ and $b$ using the given equations:
   \[
   a - b = \frac{1}{2}
   \]
   \[
   a^2 + b^2 = 1
   \]
5. Square the first equation:
   \[
   (a - b)^2 = \left(\frac{1}{2}\right)^2 \implies a^2 - 2ab + b^2 = \frac{1}{4}
   \]
6. Substitute $a^2 + b^2 = 1$ into the squared equation:
   \[
   1 - 2ab = \frac{1}{4} \implies 2ab = 1 - \frac{1}{4} = \frac{3}{4} \implies ab = \frac{3}{8}
   \]
7. Now, solve for $a$ and $b$:
   \[
   a = b + \frac{1}{2}
   \]
   Substitute into $a^2 + b^2 = 1$:
   \[
   (b + \frac{1}{2})^2 + b^2 = 1
   \]
   \[
   b^2 + b^2 + b + \frac{1}{4} = 1 \implies 2b^2 + b + \frac{1}{4} = 1
   \]
   \[
   2b^2 + b - \frac{3}{4} = 0
   \]
8. Solve the quadratic equation for $b$:
   \[
   b = \frac{-1 \pm \sqrt{1 + 6}}{4} = \frac{-1 \pm \sqrt{7}}{4}
   \]
   Since $b$ must be positive in the interval $\left(0, \frac{\pi}{2}\right)$:
   \[
   b = \frac{\sqrt{7} - 1}{4}
   \]
   Then:
   \[
   a = b + \frac{1}{2} = \frac{\sqrt{7} - 1}{4} + \frac{2}{4} = \frac{\sqrt{7} + 1}{4}
   \]
9. Now, calculate $a^3 + b^3$:
   \[
   a + b = \frac{\sqrt{7} + 1}{4} + \frac{\sqrt{7} - 1}{4} = \frac{2\sqrt{7}}{4} = \frac{\sqrt{7}}{2}
   \]
   \[
   a^2 - ab + b^2 = 1 - ab = 1 - \frac{3}{8} = \frac{5}{8}
   \]
   \[
   a^3 + b^3 = (a + b)(a^2 - ab + b^2) = \left(\frac{\sqrt{7}}{2}\right) \left(\frac{5}{8}\right) = \frac{5\sqrt{7}}{16}
   \]
10. Identify $m$, $n$, and $p$:
    \[
    m = 5, \quad n = 16, \quad p = 7
    \]
    \[
    m + n + p = 5 + 16 + 7 = 28
    \]

The final answer is $\boxed{28}$