Let $a$, $b$, and $c$ be positive real numbers such that $a^2+b^2+c^2=989$ and $(a+b)^2+(b+c)^2+(c+a)^2=2013$. Find $a+b+c$.
1. We start with the given equations:
   \[
   a^2 + b^2 + c^2 = 989
   \]
   and
   \[
   (a+b)^2 + (b+c)^2 + (c+a)^2 = 2013.
   \]

2. Expand the second equation:
   \[
   (a+b)^2 + (b+c)^2 + (c+a)^2 = a^2 + b^2 + 2ab + b^2 + c^2 + 2bc + c^2 + a^2 + 2ca.
   \]
   Simplify the expanded form:
   \[
   (a+b)^2 + (b+c)^2 + (c+a)^2 = 2(a^2 + b^2 + c^2) + 2(ab + bc + ca).
   \]

3. Substitute the given values into the simplified equation:
   \[
   2(a^2 + b^2 + c^2) + 2(ab + bc + ca) = 2013.
   \]
   Since \(a^2 + b^2 + c^2 = 989\), we substitute this into the equation:
   \[
   2(989) + 2(ab + bc + ca) = 2013.
   \]
   Simplify:
   \[
   1978 + 2(ab + bc + ca) = 2013.
   \]

4. Solve for \(ab + bc + ca\):
   \[
   2(ab + bc + ca) = 2013 - 1978,
   \]
   \[
   2(ab + bc + ca) = 35,
   \]
   \[
   ab + bc + ca = 17.5.
   \]

5. Use the identity for the square of the sum of three numbers:
   \[
   (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).
   \]
   Substitute the known values:
   \[
   (a + b + c)^2 = 989 + 2(17.5),
   \]
   \[
   (a + b + c)^2 = 989 + 35,
   \]
   \[
   (a + b + c)^2 = 1024.
   \]

6. Solve for \(a + b + c\):
   \[
   a + b + c = \sqrt{1024},
   \]
   \[
   a + b + c = 32.
   \]

The final answer is \(\boxed{32}\).