Cho a + b > 0, CMR \( a^3 + b^3 \geq a.b(a + b) \)

The problem appears to be a mathematical inequality involving \( a \) and \( b \) with the condition \( a + b > 0 \). You are asked to prove that \( a^3 + b^3 \geq ab(a + b) \).

This is a classic result in algebra that can be proven using the factorization of cubes or the concept of inequalities.

### Proof:

1. **Factorization of Cubes**:
The left-hand side \( a^3 + b^3 \) can be factored as:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
Thus, we need to show:
\[
(a + b)(a^2 - ab + b^2) \geq ab(a + b)
\]

2. **Assuming \( a + b > 0 \)**:
Since \( a + b > 0 \), we can divide both sides of the inequality by \( a + b \) (which does not change the direction of the inequality):
\[
a^2 - ab + b^2 \geq ab
\]

3. **Rearranging**:
Rearranging gives us:
\[
a^2 - 2ab + b^2 \geq 0
\]
which simplifies to:
\[
(a - b)^2 \geq 0
\]

4. **Conclusion**:
Since \( (a - b)^2 \) is always non-negative, we conclude that:
\[
a^3 + b^3 \geq ab(a + b)
\]
Thus, the inequality holds.

This completes the proof. Therefore, \( a^3 + b^3 \geq ab(a + b) \) when \( a + b > 0 \).