How to Identify a Quotient of a Given Free Group $\newcommand{\Z}{\mathbf Z}$

Problem. Let $G$ be the free group generated by three symbols $a, b$ and $c$, and let denote $G$ by writing $F(a, b, c)$. Let $N$ be the normal subgroup of $G$ generated by the words $aba^{-1}b^{-1}$ and $cbc^{-1}b^{-1}$. I want to find out what is $F(a, b, c)/N$.

I asked this of my friend and he said that his guess is
$$F(a, b, c)/N \cong \Z\times F(x, z)$$
where $F(x, z)$ denotes the free group generated by two symbols $x$ and $z$.
I am not sure how he arrived at this guess, but this seemed promising when I considered the homomorphism $F(a, b, c)\to \Z\times F(x, z)$ which satisfies
$$a\mapsto (0, x),\quad b\mapsto (1,\epsilon),\quad c\mapsto (0, z)$$
Here $\epsilon$ is the empty word in $F(x, z)$. This is surjective homomorphism whose kernel $K$ contains $aba^{-1}b^{-1}$ and $cbc^{-1}b^{-1}$. Therefore $K$ contains $N$.
So if I could show what $K=N$ then I am done. 
But I cannot see how to show this. How do I proceed?
 A: He probably arrived at the guess by noting that quotienting by $aba^{-1}b^{-1}$ means $a$ and $b$ commute.  Similarly for $c$ and $b$.  So we can pull all the $b$s out to the front, say, turning any expression into $b^nE(a,c)$ where $E(a,c)$ is some expression(/word) only involving $a$ and $c$ and $n\in\mathbb{Z}$.  At this point you can directly show the isomorphism.
One way to complete the proof in the direction you went using the above normal form is to consider what the inverse image of your homomorphism, let's call it $\varphi$, is on the identity element $(0,\epsilon)$.  You know, because $K$ contains $N$ that $\varphi(w) = (0,\epsilon)$ only when $\varphi(w') = (0,\epsilon)$ where $w' = b^nf(w)$ where $f(b) = \epsilon$,$f(a) = a$, and $f(c) = c$, i.e. $f(w)$ is $w$ with all the $b$s erased.  Now $\varphi(b^nf(w)) = (0,\epsilon)$ only when $n = 0$ and $w = \epsilon$, the latter because there is nothing relating $a$ and $c$.  This implies that the kernel of $\varphi$ is no larger than $N$.
