Given $w$ and $z$ are complex numbers such that $|w+z|=1$ and $|w^2+z^2|=14$, find the smallest possible value of $|w^3+z^3|$. Here $| \cdot |$ denotes the absolute value of a complex number, given by $|a+bi|=\sqrt{a^2+b^2}$ whenever $a$ and $b$ are real numbers.
Given the conditions \( |w+z| = 1 \) and \( |w^2+z^2| = 14 \), we need to find the smallest possible value of \( |w^3+z^3| \).

1. **Express \( w^3 + z^3 \) in terms of \( w \) and \( z \):**
   \[
   w^3 + z^3 = (w+z)(w^2 - wz + z^2)
   \]
   Given \( |w+z| = 1 \), we have:
   \[
   |w^3 + z^3| = |w+z| \cdot |w^2 - wz + z^2| = 1 \cdot |w^2 - wz + z^2| = |w^2 - wz + z^2|
   \]

2. **Use the given conditions to find \( wz \):**
   From the identity:
   \[
   (w+z)^2 = w^2 + 2wz + z^2
   \]
   and knowing \( |w+z| = 1 \), we have:
   \[
   |(w+z)^2| = |w^2 + 2wz + z^2| = 1^2 = 1
   \]
   Given \( |w^2 + z^2| = 14 \), we can write:
   \[
   |w^2 + 2wz + z^2| = |w^2 + z^2 + 2wz| = 1
   \]
   Let \( A = w^2 + z^2 \) and \( B = 2wz \). Then:
   \[
   |A + B| = 1 \quad \text{and} \quad |A| = 14
   \]
   Since \( A = w^2 + z^2 \), we have:
   \[
   |14 + 2wz| = 1
   \]
   This implies:
   \[
   14 + 2wz = \pm 1
   \]
   Solving for \( wz \):
   \[
   2wz = 1 - 14 \quad \text{or} \quad 2wz = -1 - 14
   \]
   \[
   2wz = -13 \quad \text{or} \quad 2wz = -15
   \]
   \[
   wz = -\frac{13}{2} \quad \text{or} \quad wz = -\frac{15}{2}
   \]

3. **Find \( |w^2 - wz + z^2| \):**
   Using \( wz = -\frac{13}{2} \):
   \[
   w^2 - wz + z^2 = w^2 + z^2 - wz = 14 - \left(-\frac{13}{2}\right) = 14 + \frac{13}{2} = \frac{28}{2} + \frac{13}{2} = \frac{41}{2}
   \]
   Using \( wz = -\frac{15}{2} \):
   \[
   w^2 - wz + z^2 = w^2 + z^2 - wz = 14 - \left(-\frac{15}{2}\right) = 14 + \frac{15}{2} = \frac{28}{2} + \frac{15}{2} = \frac{43}{2}
   \]

4. **Determine the smallest possible value:**
   The smallest possible value of \( |w^2 - wz + z^2| \) is:
   \[
   \min\left(\frac{41}{2}, \frac{43}{2}\right) = \frac{41}{2}
   \]

The final answer is \( \boxed{\frac{41}{2}} \).