I am trying to understand Theorem III.4.25 from the Helemskii's monograph "The Homology of Banach and Topological Algebras". This result states that
\begin{equation} \label{eq}\tag{1} \mathcal{H}_n(A,X)=Tor_n^{A^e}(X,A_+), \end{equation}
where $A$ is a Banach algebra, $X$ is an $A$-bimodule, $A_+=A\oplus\mathbb{C}$ is the unconditional unitization of $A$ and $A^e=A_+\widehat{\otimes}A_+^{op}$ is the enveloping algebra of $A$. On the left hand-side of \eqref{eq} we consider the homology groups of the complex
$\ldots\to C_{n+1}(A,X)\to C_n(A,X)\to\ldots\to C_0(A,X)\to0$,
where
$C_0(A,X)=X,\,\,C_n(A,X)=A^{\widehat{\otimes}n}\widehat{\otimes}X$
and the differentials $d_n\colon\,C_{n+1}(A,X)\to C_n(A,X)$ are given by
$d_n(a_1\otimes\ldots\otimes a_{n+1}\otimes x):=a_2\otimes\ldots\otimes a_{n+1}\otimes a_1\cdot x+\sum\limits_{k=1}^n(-1)^ka_1\otimes\ldots\otimes a_ka_{k+1}\otimes\ldots\otimes a_{n+1}\otimes x+(-1)^{n+1}a_1\otimes\ldots\otimes a_n\otimes x\cdot a_{n+1}$.
On the right hand-side of \eqref{eq} $X$ is a right $A^e$-module and $A_+$ is a left $A^e$-module and we consider the homology groups of the complex $X\widehat{\otimes}_{A^e}\mathcal{B}''(A_+)$, where $\mathcal{B}''(A_+)$ is the complex
$\ldots\to A^e_+\widehat{\otimes}(A^e)^{\widehat{\otimes}n}\widehat{\otimes}A_+\to\ldots\to A^e_+\widehat{\otimes}A^e\widehat{\otimes}A_+\to A^e_+\widehat{\otimes}A_+\to0$.
The proof that Helemskii provides suggests that this is obvious. However, since I am a complete beginner in the homology theory, I would appreciate if somebody showed me how this goes.