This problem has already been asked before but it was solved using a diagram chase. I am interested in seeing the way to do it using Long Exact Cohomology Sequence.
Let $0 \rightarrow L^{*} \rightarrow M^{*} \rightarrow N^{*} \rightarrow 0$ be a short exact sequence of cochain complexes. Prove that if any two of the complexes are exact, so is the third.
From the short exact sequence of chain complexes $$0 \to L^\bullet \to M^\bullet \to N^\bullet \to 0,$$ we obtain a long exact sequence in cohomology $$\cdots \to H^*(L^\bullet) \to H^*(M^\bullet) \to H^*(N^\bullet) \to H^{*+1}(L^\bullet) \to \cdots.$$
A cochain complex is exact iff its cohomology is zero. So if any two of $L^\bullet$, $M^\bullet$, and $N^\bullet$ are exact, we see that their corresponding cohomologies are zero. Hence in the long exact sequence, the cohomology of the third cochain complex is surrounded by zeroes and hence must be zero itself. This means that the third cochain complex is exact too.