Let $x$ and $y$ be positive real numbers. Define $a = 1 + \tfrac{x}{y}$ and $b = 1 + \tfrac{y}{x}$. If $a^2 + b^2 = 15$, compute $a^3 + b^3$.
1. Given the definitions \( a = 1 + \frac{x}{y} \) and \( b = 1 + \frac{y}{x} \), we start by noting that \( a \) and \( b \) are positive real numbers.
2. We are given that \( a^2 + b^2 = 15 \).
3. We also note that \( \frac{1}{a} + \frac{1}{b} = 1 \). This can be derived as follows:
   \[
   \frac{1}{a} = \frac{1}{1 + \frac{x}{y}} = \frac{y}{x + y}
   \]
   \[
   \frac{1}{b} = \frac{1}{1 + \frac{y}{x}} = \frac{x}{x + y}
   \]
   Adding these two fractions:
   \[
   \frac{1}{a} + \frac{1}{b} = \frac{y}{x + y} + \frac{x}{x + y} = \frac{y + x}{x + y} = 1
   \]
4. Let \( a + b = c \) and \( ab = d \). We know from the identity for the sum of squares:
   \[
   a^2 + b^2 = (a + b)^2 - 2ab
   \]
   Substituting the given \( a^2 + b^2 = 15 \):
   \[
   15 = c^2 - 2d
   \]
5. From the identity \( \frac{1}{a} + \frac{1}{b} = 1 \), we have:
   \[
   \frac{a + b}{ab} = 1 \implies \frac{c}{d} = 1 \implies c = d
   \]
6. Substituting \( d = c \) into the equation \( 15 = c^2 - 2d \):
   \[
   15 = c^2 - 2c
   \]
   \[
   c^2 - 2c - 15 = 0
   \]
7. Solving this quadratic equation:
   \[
   c = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm 8}{2}
   \]
   \[
   c = 5 \quad \text{or} \quad c = -3
   \]
   Since \( a \) and \( b \) are positive, \( c \) must be positive, so \( c = 5 \).
8. Therefore, \( a + b = 5 \) and \( ab = 5 \).
9. To find \( a^3 + b^3 \), we use the identity:
   \[
   a^3 + b^3 = (a + b)((a + b)^2 - 3ab)
   \]
   Substituting \( a + b = 5 \) and \( ab = 5 \):
   \[
   a^3 + b^3 = 5((5)^2 - 3 \cdot 5) = 5(25 - 15) = 5 \cdot 10 = 50
   \]

The final answer is \(\boxed{50}\)