Find all pairs of solutions $(x,y)$:

\[ x^3 + x^2y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1). \]
1. Given the equation:
   \[
   x^3 + x^2y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1)
   \]
   we start by introducing the symmetric sums \( s = x + y \) and \( p = xy \).

2. Rewrite the given equation in terms of \( s \) and \( p \):
   \[
   x^3 + y^3 + x^2y + xy^2 = 8(x^2 + y^2 + xy + 1)
   \]
   Using the identities for symmetric sums, we have:
   \[
   x^3 + y^3 = (x+y)(x^2 - xy + y^2) = s(s^2 - 3p)
   \]
   and
   \[
   x^2 + y^2 = s^2 - 2p
   \]
   Substituting these into the equation, we get:
   \[
   s(s^2 - 3p) + p(x+y) = 8(s^2 - 2p + p + 1)
   \]
   Simplifying, we have:
   \[
   s(s^2 - 3p) + ps = 8(s^2 - p + 1)
   \]
   \[
   s^3 - 3sp + ps = 8s^2 - 8p + 8
   \]
   \[
   s^3 - 2sp = 8s^2 - 8p + 8
   \]

3. Rearrange the equation to isolate terms involving \( p \):
   \[
   s^3 - 8s^2 + 8 = 2sp - 8p
   \]
   Factor out \( p \) on the right-hand side:
   \[
   s^3 - 8s^2 + 8 = 2p(s - 4)
   \]

4. Notice that \( s - 4 \) must divide the left-hand side:
   \[
   s - 4 \mid s^3 - 8s^2 + 8
   \]
   Perform polynomial division or use the Remainder Theorem to find the possible values of \( s \). Since \( s \) must be even, we check the divisors of 56 (as \( s - 4 \mid 56 \)).

5. Possible values for \( s \) are \( 4, 6, 8, 10, 12 \). We check each value to find corresponding \( p \) and then solve for \( x \) and \( y \).

6. For \( s = 4 \):
   \[
   4^3 - 8 \cdot 4^2 + 8 = 2p(4 - 4)
   \]
   \[
   64 - 128 + 8 = 0
   \]
   This is not possible since it leads to a contradiction.

7. For \( s = 6 \):
   \[
   6^3 - 8 \cdot 6^2 + 8 = 2p(6 - 4)
   \]
   \[
   216 - 288 + 8 = 4p
   \]
   \[
   -64 = 4p \implies p = -16
   \]
   Solving \( t^2 - 6t - 16 = 0 \):
   \[
   t = \frac{6 \pm \sqrt{36 + 64}}{2} = \frac{6 \pm 10}{2}
   \]
   \[
   t = 8 \text{ or } -2
   \]
   Thus, \( (x, y) = (8, -2) \) or \( (-2, 8) \).

8. For \( s = 8 \):
   \[
   8^3 - 8 \cdot 8^2 + 8 = 2p(8 - 4)
   \]
   \[
   512 - 512 + 8 = 8p
   \]
   \[
   8 = 8p \implies p = 1
   \]
   Solving \( t^2 - 8t + 1 = 0 \):
   \[
   t = \frac{8 \pm \sqrt{64 - 4}}{2} = \frac{8 \pm \sqrt{60}}{2}
   \]
   \[
   t = 4 \pm \sqrt{15}
   \]
   Thus, \( (x, y) = (4 + \sqrt{15}, 4 - \sqrt{15}) \) or \( (4 - \sqrt{15}, 4 + \sqrt{15}) \).

9. For \( s = 10 \):
   \[
   10^3 - 8 \cdot 10^2 + 8 = 2p(10 - 4)
   \]
   \[
   1000 - 800 + 8 = 12p
   \]
   \[
   208 = 12p \implies p = \frac{208}{12} = \frac{52}{3}
   \]
   This is not an integer, so no solutions for \( s = 10 \).

10. For \( s = 12 \):
    \[
    12^3 - 8 \cdot 12^2 + 8 = 2p(12 - 4)
    \]
    \[
    1728 - 1152 + 8 = 16p
    \]
    \[
    584 = 16p \implies p = 36.5
    \]
    This is not an integer, so no solutions for \( s = 12 \).

The final answer is \(\boxed{(8, -2), (-2, 8), (4 + \sqrt{15}, 4 - \sqrt{15}), (4 - \sqrt{15}, 4 + \sqrt{15})}\).