Are there any real numbers $a,b,c$ such that $a+b+c=6$, $ab+bc+ca=9$ and $a^4+b^4+c^4=260$? What about if we let $a^4+b^4+c^4=210$?

[i] (Andrei Bâra)[/i]
To determine if there are any real numbers \(a, b, c\) such that \(a + b + c = 6\), \(ab + bc + ca = 9\), and \(a^4 + b^4 + c^4 = 260\), we can use the symmetric polynomial identities and properties of roots of polynomials.

1. **Express \(a^4 + b^4 + c^4\) in terms of symmetric polynomials:**

   Given:
   \[
   a + b + c = 6, \quad ab + bc + ca = 9
   \]
   Let \(p = a + b + c\), \(q = ab + bc + ca\), and \(r = abc\). We know:
   \[
   p = 6, \quad q = 9
   \]

2. **Use the identity for the sum of fourth powers:**

   The identity for the sum of fourth powers of the roots of a polynomial is:
   \[
   a^4 + b^4 + c^4 = (a + b + c)^4 - 4(a + b + c)^2(ab + bc + ca) + 2(ab + bc + ca)^2 + 4abc(a + b + c)
   \]
   Substituting \(p\) and \(q\):
   \[
   a^4 + b^4 + c^4 = 6^4 - 4 \cdot 6^2 \cdot 9 + 2 \cdot 9^2 + 4 \cdot 6 \cdot abc
   \]
   Simplify:
   \[
   a^4 + b^4 + c^4 = 1296 - 4 \cdot 36 \cdot 9 + 2 \cdot 81 + 24 \cdot abc
   \]
   \[
   a^4 + b^4 + c^4 = 1296 - 1296 + 162 + 24abc
   \]
   \[
   a^4 + b^4 + c^4 = 162 + 24abc
   \]

3. **Solve for \(abc\) when \(a^4 + b^4 + c^4 = 260\):**

   Given \(a^4 + b^4 + c^4 = 260\):
   \[
   260 = 162 + 24abc
   \]
   \[
   98 = 24abc
   \]
   \[
   abc = \frac{98}{24} = \frac{49}{12}
   \]

4. **Check the discriminant of the cubic polynomial:**

   The roots \(a, b, c\) are the roots of the polynomial:
   \[
   t^3 - pt^2 + qt - r = 0
   \]
   Substituting \(p, q, r\):
   \[
   t^3 - 6t^2 + 9t - \frac{49}{12} = 0
   \]
   The discriminant \(\Delta\) of this cubic polynomial must be non-negative for real roots to exist. The discriminant of a cubic polynomial \(t^3 + at^2 + bt + c\) is given by:
   \[
   \Delta = 18abc - 4b^3 + b^2a^2 - 4ac^3 - 27c^2
   \]
   Substituting \(a = -6\), \(b = 9\), and \(c = -\frac{49}{12}\):
   \[
   \Delta = 18(-6)(9)\left(-\frac{49}{12}\right) - 4(9)^3 + (9)^2(-6)^2 - 4(-6)\left(-\frac{49}{12}\right)^3 - 27\left(-\frac{49}{12}\right)^2
   \]
   Simplifying this expression is complex, but it can be shown that \(\Delta < 0\), indicating no real roots.

5. **Conclusion for \(a^4 + b^4 + c^4 = 260\):**

   Since the discriminant is negative, there are no real numbers \(a, b, c\) satisfying the given conditions.

6. **For \(a^4 + b^4 + c^4 = 210\):**

   Similarly, we solve:
   \[
   210 = 162 + 24abc
   \]
   \[
   48 = 24abc
   \]
   \[
   abc = 2
   \]

   The polynomial becomes:
   \[
   t^3 - 6t^2 + 9t - 2 = 0
   \]

   The discriminant of this polynomial can be checked similarly, and it can be shown that it has real roots. One possible set of roots is \((2, 2 - \sqrt{3}, 2 + \sqrt{3})\).

The final answer is \( \boxed{ (2, 2 - \sqrt{3}, 2 + \sqrt{3}) } \) and permutations for \(a^4 + b^4 + c^4 = 210\). There are no real solutions for \(a^4 + b^4 + c^4 = 260\).