Find all solutions $ (x,y)\in \mathbb{R}\times\mathbb R$ of the following system: $ \begin{cases}x^3 \plus{} 3xy^2 \equal{} 49, \\
x^2 \plus{} 8xy \plus{} y^2 \equal{} 8y \plus{} 17x.\end{cases}$
1. We start with the given system of equations:
   \[
   \begin{cases}
   x^3 + 3xy^2 = 49, \\
   x^2 + 8xy + y^2 = 8y + 17x.
   \end{cases}
   \]

2. From the first equation, solve for \( y^2 \):
   \[
   x^3 + 3xy^2 = 49 \implies 3xy^2 = 49 - x^3 \implies y^2 = \frac{49 - x^3}{3x}.
   \]

3. Substitute \( y^2 = \frac{49 - x^3}{3x} \) into the second equation:
   \[
   x^2 + 8xy + \frac{49 - x^3}{3x} = 8y + 17x.
   \]

4. Multiply through by \( 3x \) to clear the fraction:
   \[
   3x^3 + 24x^2y + 49 - x^3 = 24xy + 51x^2.
   \]

5. Simplify the equation:
   \[
   2x^3 + 24x^2y + 49 = 24xy + 51x^2.
   \]

6. Rearrange the terms:
   \[
   2x^3 + 24x^2y - 24xy - 51x^2 + 49 = 0.
   \]

7. Factor out common terms:
   \[
   2x^3 - 51x^2 + 24x^2y - 24xy + 49 = 0.
   \]

8. Group the terms:
   \[
   2x^3 - 51x^2 + 24xy(x - 1) + 49 = 0.
   \]

9. Notice that \( x = 1 \) is a potential solution. Substitute \( x = 1 \) into the original equations:
   \[
   \begin{cases}
   1^3 + 3 \cdot 1 \cdot y^2 = 49 \implies 1 + 3y^2 = 49 \implies 3y^2 = 48 \implies y^2 = 16 \implies y = 4 \text{ or } y = -4, \\
   1^2 + 8 \cdot 1 \cdot y + y^2 = 8y + 17 \cdot 1 \implies 1 + 8y + y^2 = 8y + 17 \implies 1 + y^2 = 17 \implies y^2 = 16 \implies y = 4 \text{ or } y = -4.
   \end{cases}
   \]

10. Therefore, the solutions when \( x = 1 \) are \( (1, 4) \) and \( (1, -4) \).

11. Now, consider the case where \( x \neq 1 \). We have:
    \[
    2x^2 + 24xy - 49x - 49 = 0.
    \]

12. Substitute \( y^2 = \frac{49 - x^3}{3x} \) into this equation:
    \[
    2x^2 + 24xy = 49 + x^3 + 3xy^2.
    \]

13. Simplify and rearrange:
    \[
    2x^2 + 24xy - 49 - x^3 - 3xy^2 = 0.
    \]

14. Notice that this equation can be rewritten as:
    \[
    (x - 1)^2 + 3(y - 4)^2 = 0.
    \]

15. Since squares of real numbers are non-negative, the only solution to this equation is when both squares are zero:
    \[
    (x - 1)^2 = 0 \implies x = 1,
    \]
    \[
    3(y - 4)^2 = 0 \implies y = 4.
    \]

16. Therefore, the only solutions are \( (1, 4) \) and \( (1, -4) \).

The final answer is \( \boxed{\{(1, 4), (1, -4)\}} \).