Determine an equation of third degree with integral coefficients having roots $\sin \frac{\pi}{14}, \sin \frac{5 \pi}{14}$ and $\sin \frac{-3 \pi}{14}.$
1. We start by noting that the roots of the polynomial we seek are $\sin \frac{\pi}{14}$, $\sin \frac{5 \pi}{14}$, and $\sin \frac{-3 \pi}{14}$. Using the identity $\sin(-x) = -\sin(x)$, we can rewrite $\sin \frac{-3 \pi}{14}$ as $-\sin \frac{3 \pi}{14}$.

2. We use the trigonometric identity for cosine of multiple angles:
   \[
   \cos(7x) = 64\cos^7(x) - 112\cos^5(x) + 56\cos^3(x) - 7\cos(x)
   \]
   This identity can be transformed into a polynomial equation in terms of $\cos(x)$.

3. We also use the identity for cosine in terms of sine:
   \[
   \cos(x) = \sin\left(\frac{\pi}{2} - x\right)
   \]
   This allows us to relate the sines of the given angles to cosines of other angles.

4. We know that:
   \[
   \sin \frac{\pi}{14} = \cos \left(\frac{3\pi}{7}\right), \quad \sin \frac{5\pi}{14} = \cos \left(\frac{\pi}{7}\right), \quad \text{and} \quad \sin \frac{-3\pi}{14} = -\sin \frac{3\pi}{14} = -\cos \left(\frac{5\pi}{7}\right)
   \]

5. The polynomial whose roots are $\cos \frac{\pi}{7}$, $\cos \frac{3\pi}{7}$, and $\cos \frac{5\pi}{7}$ is given by:
   \[
   8x^3 - 4x^2 - 4x - 1 = 0
   \]

6. Since $\sin \frac{\pi}{14}$, $\sin \frac{5\pi}{14}$, and $\sin \frac{-3\pi}{14}$ correspond to $\cos \frac{3\pi}{7}$, $\cos \frac{\pi}{7}$, and $\cos \frac{5\pi}{7}$ respectively, the polynomial with these roots is the same as the one above.

Therefore, the polynomial with integral coefficients having roots $\sin \frac{\pi}{14}$, $\sin \frac{5\pi}{14}$, and $\sin \frac{-3\pi}{14}$ is:
\[
\boxed{8x^3 - 4x^2 - 4x - 1 = 0}
\]