Weibel HA Exercise 1.5.9 I cannot solve the following problem from Weibel:

Let $f:B\to C$ be a map of chain complexes. Show that the natural maps $\alpha : \ker(f)[-1]\to \operatorname{cone}(f)$ and $\beta: \operatorname{cone}(f)\to \operatorname{coker}(f)$ give rise to a long exact sequence:
  $$\cdots \to H_{n-1}(\ker(f))\to H_n(\operatorname{cone}(f))\to H_n(\operatorname{coker}(f))\to H_{n-2}(\ker(f)) \to \cdots$$

Any help is appreciated!
 A: Here's a sketch. The details it leaves to be checked are fairly straightforward, I hope.
Let $g:\text{im}(f)\to C$ be the inclusion map.
There is a short exact sequence of complexes
$$0\to\ker(f)[-1]\to\text{cone}(f)\to\text{cone}(g)\to0,$$
which gives a long exact sequence of homology.
The natural map $\text{cone}(f)\to\text{cok}(f)$ factors through $\text{cone}(g)$, with the resulting map $\text{cone}(g)\to\text{cok}(f)$ a quasi-isomorphism.
A: I couldn't figure it out, and here is more an idea or comment rather than an answer, but it was too long to leave as a comment. At most I can  define the connecting map $\delta: H_{n}(coker(f)) \to H_{n-2}(ker(f))$ 
Define the complex $D=im(f)$ and so $D\cong B/ker(f)$.
Then we have short exact sequences 
$\begin{align}
0 \to &ker(f)[-1] \to B[-1] \to D[-1] \to 0 \\
0 \to C \to &cone(f) \to B[-1] \to 0 \\
0 \to D \to C \to &coker(f) \to 0
\end{align}$
and there are also a few obvious vertical arrows (e.g. $C=C$ and $B[-1] = B[-1]$) between these rows giving a commutative diagram.
Then taking the long exact sequence for each of the rows and hope they are tied together nicely.
Also define the connecting map $\delta: H_{n}(coker(f)) \to H_{n-2}(ker(f))$ in the exercise as the composite of $H_n(coker(f)) \to H_{n-1}(D)$ (connecting morphism of the third row)  and $H_{n-1}D \to H_{n-2}(ker f)$ (connecting morphism of first row). 
Now let's try (and fail) to check exactness at $H_n(cone(f))$. Let $x \in H_n(cone(f))$ be such that $\beta_*x =0 \in H_n(coker(f))$. Push $x$ to $x' \in H_{n-1}(B)$ using the second row. Then push $x'$ to $x'' \in H_{n-1}(D)$ using the first row. Using the connecting arrow of the third row push $\beta_* x =0 $ to $0 \in H_{n-1}D$. If this means $x'' =0$ (it is not clear that it does), then by exactness we get $X \in H_{n-1}(ker(f))$ mapping to $x' \in H_{n-1}B$. Its not clear to me that $X$ maps to $x$ though. :(
