Writing specific word as product of commutators in free groups Let $F$ be a free group on $\{a,b,c,\cdots\}$. Then a word 
$$a^mb^nc^k\cdots \hskip1cm \mbox{(finite expression)}$$
lies in commutator subgroup $[F,F]$ if and only if sum of powers of $a$ is zero, sum of powers of $b$ is zero, and so on. This follows by considering the natural homomorphism from $F$ to $F/[F,F]$. 
Now, my question, once we decide that a certain word in $F$ is in $[F,F]$, how to write it as a product of commutators? Is there any inductive procedure?
For example, $a^2bc^{-2}a^{-2}b^{-1}c^2.$ This is in $[F,F]$. Is there an algorithm, or inductive procedure, to write this word as product of commutators? (Further examples would be better).
 A: This is only a partial answer but it is too long for a comment.
A long commutator of $G$ is an element $g \in G$ of which can be expressed in the form
$$a_1^{-1} a_2^{-1} \cdots a_n^{-1} a_1 a_2 \cdots a_n, \hspace{2ex} a_i \in G.$$
We call $n$ the length of the expression of $g$.
Fact:(1) Any product of $k$ commutators is a long commutator which has a realization of length $2k$.
(2) Any long commutator with a realization of length $m$, where $m = 2k$ or $m = 2k+1$, is a product of $k$ commutators.
Proof of (2): Since $a_1^{-1} a_2^{-1} a_1 a_2$ and $a_1^{-1} a_2^{-1} a_3^{-1} a_1 a_2 a_3 = (a_2 a_1)^{-1} (a_2 a_3)^{-1} (a_2 a_1) (a_2 a_3)$ are both long commutators, the statement is true for $m = 2$ and $m = 3$.
Suppose it is true when $m = r$. Then
$$a_1^{-1} \cdots a_{r+2}^{-1} a_1 \cdots a_{r+2} =$$
$$= a_1^{-1} \cdots a_r^{-1} a_1 \cdots a_r (a_{r+1} a_1 \cdots a_r)^{-1} (a_{r+1} a_{r+2})^{-1} (a_{r+1} a_1 \cdots a_r) (a_{r+1} a_{r+2}).
$$
By the inductive hypothesis, the expression $a_1^{-1} \cdots a_r^{-1} a_1 \cdots a_r$ may be written as $\frac{1}{2} r$ or $\frac{1}{2}(r- 1)$ commutators, according as $r$ is even or odd. The remaining expressions constitute a single commutator which proves the assertion.
So for any long commutator this proof gives an inductive procedure to produce a product of long commutators. Your example is one of length $3$ (see induction start of the proof). So it suffices to find a procedure to transform your given element of $[F, F]$ into a long commutator which may be easier.
A: In Applications of topological graph theory to group theory, Gorenstein and Turner give such an algorithm. Furthermore, the number of commutators in the product they find is minimal.
I give some details on my blog, and in particular I show how the argument gives $$x^2yx^{-1}y^{-2}x^{-1}y=[x^2yx^{-1},y^{-1}x^{-1}].$$
