Let $A$ be a $k$-algebra, $k$ a commutative ring and $M$ an $A$ bimodule. Suppose that both $A$ and $M$ are projective as modules over $k$. Why then $A\otimes M \otimes A$ is projective as left $A^e := A \otimes A^{op}$ module? Here all tensors are taken over $k$ and action is given by $$A^e \times A\otimes M \otimes A \to A\otimes M \otimes A \\ (a\otimes b, a_1\otimes m \otimes a_2) \to aa_1\otimes m \otimes a_2 b$$
Attempt 1: There is a free $k$-module $F$ and $k$-module $M'$ such that $M\oplus M'=F$. By tensoring under $k$ on left and on the right, one gets $A \otimes M \otimes A \oplus A\otimes M' \otimes A = A\otimes F \otimes A$... If I could justify that $A\otimes F \otimes A$ is free as $A^e$-module, I'm done. However, I'am aware that I didn't use the projectivity of $A$.
Attempt 2: It suffices to show that functor $Hom_{A^e} (A\otimes M \otimes A,-) $ is exact. By adjoint functior theorem for modules this functor is isomorphic to $Hom_k(A, Hom_k(M, Hom_{A^e}(A,-)))$, but I still can't conclude because of that mysterious $Hom_{A^e} (A,-)$.