Solve the system $\begin{cases} x^3 - y^3 = 26 \\
x^2y - xy^2 = 6  
\end{cases}$

[hide=other version]solved below
Solve the system $\begin{cases} x^3 - y^3 = 2b \\
x^2y - xy^2 = b  
\end{cases}$[/hide]
We are given the system of equations:
\[
\begin{cases}
x^3 - y^3 = 2b \\
x^2y - xy^2 = b
\end{cases}
\]

1. First, we use the identity for the difference of cubes:
\[
x^3 - y^3 = (x-y)(x^2 + xy + y^2)
\]
So, the first equation becomes:
\[
(x-y)(x^2 + xy + y^2) = 2b \tag{1}
\]

2. For the second equation, we factor out \(xy\):
\[
x^2y - xy^2 = xy(x - y) = b \tag{2}
\]

3. Now, we divide equation (1) by equation (2):
\[
\frac{(x-y)(x^2 + xy + y^2)}{xy(x-y)} = \frac{2b}{b}
\]
Simplifying, we get:
\[
\frac{x^2 + xy + y^2}{xy} = 2
\]
This simplifies to:
\[
\frac{x^2}{xy} + \frac{xy}{xy} + \frac{y^2}{xy} = 2
\]
\[
\frac{x}{y} + 1 + \frac{y}{x} = 2
\]
Let \( t = \frac{x}{y} \), then:
\[
t + 1 + \frac{1}{t} = 2
\]
Multiplying through by \(t\), we get:
\[
t^2 + t + 1 = 2t
\]
\[
t^2 - t + 1 = 0
\]

4. Solving the quadratic equation \(t^2 - t + 1 = 0\):
\[
t = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} = \frac{1 \pm i\sqrt{3}}{2}
\]
So, \( t = \frac{1 + i\sqrt{3}}{2} \) or \( t = \frac{1 - i\sqrt{3}}{2} \).

5. Since \( t = \frac{x}{y} \), we have:
\[
x = y \cdot \frac{1 + i\sqrt{3}}{2} \quad \text{or} \quad x = y \cdot \frac{1 - i\sqrt{3}}{2}
\]

6. We also need to consider the case when \( x = y \):
\[
x^3 - y^3 = 2b \implies 0 = 2b \implies b = 0
\]
If \( b = 0 \), then the second equation \( x^2y - xy^2 = 0 \) is always true for \( x = y \).

7. For \( b \neq 0 \), we use the solutions for \( t \):
\[
x = y \cdot \frac{1 + i\sqrt{3}}{2} \quad \text{or} \quad x = y \cdot \frac{1 - i\sqrt{3}}{2}
\]

8. Substituting back into the original equations, we verify that these solutions satisfy both equations.

Therefore, the solutions to the system are:
\[
(x, y) = (c, c) \quad \text{for} \quad b = 0
\]
\[
(x, y) = \left( y \cdot \frac{1 + i\sqrt{3}}{2}, y \right) \quad \text{or} \quad \left( y \cdot \frac{1 - i\sqrt{3}}{2}, y \right) \quad \text{for} \quad b \neq 0
\]

The final answer is \( \boxed{ (x, y) = (c, c) } \) for \(b = 0\) and \((x, y) = \left( y \cdot \frac{1 + i\sqrt{3}}{2}, y \right)\) or \(\left( y \cdot \frac{1 - i\sqrt{3}}{2}, y \right)\) for \(b \neq 0\).