If $A$ and $B$ are relatively prime numbers, then the greatest common divisor of $A^{3}+B^{3}$ and $A^{2}+B^{2}$ is 1 or 2.
(1) Suppose that $\left(A^{2}+B^{2}\right)$ and $(A+B)$ have a common divisor; then this common divisor is also a divisor of $\left(A^{2}+B^{2}\right)-\left(A^{2}+B^{2}\right)=2 A B$; but one of the divisors of $2 A B$ is 2, which is a divisor of $\left(A^{2}+B^{2}\right)$ and $(A+B)$ only if $A$ and $B$ are odd numbers. $2 A B$ has another divisor which is either a divisor of $A$ or $B$, for example, $A$.

But if $A$ and $(A+B)$ have a common divisor, then this divisor is also a divisor of $B$, which is not possible according to the condition, and thus $\left(A^{2}+B^{2}\right)$ and $(A+B)$ have no other common divisor except 1 and 2.

(2) If $\left(A^{2}+B^{2}\right)$ and $\left(A^{2}-A B+B^{2}\right)$ have a common divisor, then this common divisor is also a divisor of $\left(A^{2}+B^{2}\right)-\left(A^{2}-A B+B^{2}\right)=A B$. But the common divisor of $\left(A^{2}+B^{2}\right)$ and $A B$ is also a divisor of one of the factors of the product $A B$, for example, $A$. But if $A$ and $\left(A^{2}+B^{2}\right)$ have a common divisor, then this divisor is also a divisor of $B^{2}$, and thus of $B$; which contradicts the condition, so $\left(A^{2}+B^{2}\right)$ and $\left(A^{2}-A B+B^{2}\right)$ can have no other common divisor except 1.

From (1) and (2) it follows that the greatest common divisor of $\left(A^{2}+B^{2}\right)$ and $(A+B)\left(A^{2}-A B+B^{2}\right)=\left(A^{3}+B^{3}\right)$ is 1 or 2.