Find all pairs $(x, y)$ of real numbers satisfying the system :
$\begin{cases} x + y = 3 \\
x^4 - y^4 = 8x - y \end{cases}$
1. Start with the given system of equations:
   \[
   \begin{cases}
   x + y = 3 \\
   x^4 - y^4 = 8x - y
   \end{cases}
   \]

2. From the first equation, solve for \( y \):
   \[
   y = 3 - x
   \]

3. Substitute \( y = 3 - x \) into the second equation:
   \[
   x^4 - (3 - x)^4 = 8x - (3 - x)
   \]

4. Expand \( (3 - x)^4 \):
   \[
   (3 - x)^4 = 81 - 108x + 54x^2 - 12x^3 + x^4
   \]

5. Substitute the expanded form back into the equation:
   \[
   x^4 - (81 - 108x + 54x^2 - 12x^3 + x^4) = 8x - 3 + x
   \]

6. Simplify the equation:
   \[
   x^4 - 81 + 108x - 54x^2 + 12x^3 - x^4 = 9x - 3
   \]
   \[
   12x^3 - 54x^2 + 108x - 81 = 9x - 3
   \]

7. Combine like terms:
   \[
   12x^3 - 54x^2 + 99x - 78 = 0
   \]

8. Factor out the common term:
   \[
   3(4x^3 - 18x^2 + 33x - 26) = 0
   \]

9. Solve the cubic equation \( 4x^3 - 18x^2 + 33x - 26 = 0 \). Using the Rational Root Theorem, we test possible rational roots and find that \( x = 2 \) is a root.

10. Perform polynomial division to factorize the cubic polynomial:
    \[
    4x^3 - 18x^2 + 33x - 26 = (x - 2)(4x^2 - 10x + 13)
    \]

11. Solve the quadratic equation \( 4x^2 - 10x + 13 = 0 \) using the quadratic formula:
    \[
    x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
    \]
    where \( a = 4 \), \( b = -10 \), and \( c = 13 \):
    \[
    x = \frac{10 \pm \sqrt{100 - 208}}{8} = \frac{10 \pm \sqrt{-108}}{8}
    \]
    \[
    x = \frac{10 \pm \sqrt{108}i}{8} = \frac{10 \pm 6\sqrt{3}i}{8} = \frac{5 \pm 3\sqrt{3}i}{4}
    \]
    Since the roots are complex, the only real solution is \( x = 2 \).

12. Substitute \( x = 2 \) back into \( y = 3 - x \):
    \[
    y = 3 - 2 = 1
    \]

13. Therefore, the only real pair \((x, y)\) is:
    \[
    \boxed{(2, 1)}
    \]