Find all functions $f\colon \mathbb{R}\to\mathbb{R}$ that satisfy  $f(x+y)-f(x-y)=2y(3x^2+y^2)$ for all $x,y{\in}R$


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Azerbaijan Land of the Fire :lol:
1. Let \( P(x, y) \) be the assertion \( f(x+y) - f(x-y) = 2y(3x^2 + y^2) \).

2. First, we test the function \( f(x) = x^3 + a \) to see if it satisfies the given functional equation. Assume \( f(x) = x^3 + a \).

3. Substitute \( f(x) = x^3 + a \) into the functional equation:
   \[
   f(x+y) = (x+y)^3 + a
   \]
   \[
   f(x-y) = (x-y)^3 + a
   \]

4. Calculate \( f(x+y) - f(x-y) \):
   \[
   f(x+y) - f(x-y) = (x+y)^3 + a - ((x-y)^3 + a)
   \]
   \[
   = (x+y)^3 - (x-y)^3
   \]

5. Expand the cubes:
   \[
   (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
   \]
   \[
   (x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3
   \]

6. Subtract the expanded forms:
   \[
   (x+y)^3 - (x-y)^3 = (x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3)
   \]
   \[
   = x^3 + 3x^2y + 3xy^2 + y^3 - x^3 + 3x^2y - 3xy^2 + y^3
   \]
   \[
   = 6x^2y + 2y^3
   \]

7. Compare this with the right-hand side of the original equation:
   \[
   2y(3x^2 + y^2)
   \]
   \[
   = 6x^2y + 2y^3
   \]

8. Since both sides match, \( f(x) = x^3 + a \) is indeed a solution.

9. To show that this is the only solution, consider the general form of the functional equation. Suppose \( f \) is any function satisfying the given equation. We need to show that \( f(x) \) must be of the form \( x^3 + a \).

10. Let \( x = 0 \) in the original equation:
    \[
    f(y) - f(-y) = 2y(y^2)
    \]
    \[
    f(y) - f(-y) = 2y^3
    \]

11. This implies that \( f(y) = f(-y) + 2y^3 \). Let \( f(y) = g(y) + y^3 \), where \( g(y) \) is an even function (i.e., \( g(y) = g(-y) \)).

12. Substitute \( f(x) = g(x) + x^3 \) into the original equation:
    \[
    g(x+y) + (x+y)^3 - g(x-y) - (x-y)^3 = 2y(3x^2 + y^2)
    \]
    \[
    g(x+y) - g(x-y) + (x+y)^3 - (x-y)^3 = 2y(3x^2 + y^2)
    \]

13. From the previous steps, we know that \( (x+y)^3 - (x-y)^3 = 6x^2y + 2y^3 \). Therefore:
    \[
    g(x+y) - g(x-y) + 6x^2y + 2y^3 = 2y(3x^2 + y^2)
    \]
    \[
    g(x+y) - g(x-y) = 0
    \]

14. Since \( g(x+y) = g(x-y) \) for all \( x \) and \( y \), and \( g \) is an even function, it must be constant. Let \( g(x) = c \).

15. Therefore, \( f(x) = x^3 + c \).

16. Let \( c = a \), where \( a \) is a constant. Thus, the general solution is \( f(x) = x^3 + a \).

\(\blacksquare\)

The final answer is \( \boxed{ f(x) = x^3 + a } \) where \( a \in \mathbb{R} \).