Real numbers $a$ and $b$ satisfy the system of equations $$\begin{cases} a^3-a^2+a-5=0 \\ b^3-2b^2+2b+4=0  \end{cases}$$  Find the numerical value of the sum $a+ b$.
Given the system of equations:
\[
\begin{cases}
a^3 - a^2 + a - 5 = 0 \\
b^3 - 2b^2 + 2b + 4 = 0
\end{cases}
\]
we need to find the numerical value of the sum \(a + b\).

1. Let \(a + b = k\). Then \(b = k - a\).

2. Substitute \(b = k - a\) into the second equation:
\[
(k - a)^3 - 2(k - a)^2 + 2(k - a) + 4 = 0
\]

3. Expand and simplify the equation:
\[
(k - a)^3 = k^3 - 3k^2a + 3ka^2 - a^3
\]
\[
-2(k - a)^2 = -2(k^2 - 2ka + a^2) = -2k^2 + 4ka - 2a^2
\]
\[
2(k - a) = 2k - 2a
\]
\[
\Rightarrow k^3 - 3k^2a + 3ka^2 - a^3 - 2k^2 + 4ka - 2a^2 + 2k - 2a + 4 = 0
\]

4. Substitute \(a^3\) from the first equation \(a^3 = a^2 - a + 5\):
\[
k^3 - 3k^2a + 3ka^2 - (a^2 - a + 5) - 2k^2 + 4ka - 2a^2 + 2k - 2a + 4 = 0
\]
\[
k^3 - 3k^2a + 3ka^2 - a^2 + a - 5 - 2k^2 + 4ka - 2a^2 + 2k - 2a + 4 = 0
\]

5. Combine like terms:
\[
k^3 - 3k^2a - 2k^2 + 3ka^2 - a^2 - 2a^2 + a + 4ka - 2a - 5 + 2k + 4 = 0
\]
\[
k^3 - (3k^2 + 2k^2)a + (3ka^2 - a^2 - 2a^2) + (a + 4ka - 2a) + (2k - 5 + 4) = 0
\]
\[
k^3 - 5k^2a + ka^2 + 2ka - a - 1 = 0
\]

6. Factor out common terms:
\[
k^3 - 5k^2a + (k - 1)a^2 + 2ka - a - 1 = 0
\]

7. Since \(a\) and \(b\) are real numbers, the discriminant of the quadratic equation in \(a\) must be non-negative. However, the discriminant \(\Delta\) of the quadratic equation \(3a^2 - (3k - 1)a + (k^2 - k + 1) = 0\) is:
\[
\Delta = (3k - 1)^2 - 4 \cdot 3 \cdot (k^2 - k + 1)
\]
\[
\Delta = 9k^2 - 6k + 1 - 12k^2 + 12k - 12
\]
\[
\Delta = -3k^2 + 6k - 11
\]

8. For \(\Delta\) to be non-negative, we need:
\[
-3(k - 1)^2 - 8 < 0
\]
This is always true for real \(k\), thus we have a contradiction unless \(k = 1\).

9. Therefore, \(a + b = k = 1\).

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