The given expression can also be written as:

$$
\begin{gathered}
{\left[\left(\frac{x+y}{a-b}\right)^{3}\right]^{2}\left[\left(\frac{b-a}{x-y}\right)^{2}\right]^{3}\left[\left(\frac{x-y}{x+y}\right)^{2}\right]^{3}=} \\
\left(\frac{x+y}{a-b}\right)^{6}\left(\frac{b-a}{x-y}\right)^{6}\left(\frac{x-y}{x+y}\right)^{6}=\left(\frac{b-a}{a-b}\right)^{6}=(-1)^{6}=1
\end{gathered}
$$

(Woltitz Renée, Budapest, VI. district.)

The problem was also solved by: Dévai E., Eckstein V., Ehrenstein P., Freund E. Füstös P., Harsányi Z., Heimlich P., Jánosy Gy., Kiss J., Makó E., Paunz A., Pichler S., Sárközy E., Schlesinger Gy., Schwarz Gy., Schuster Gy., Szabó J., Szécsi I., Tóth B.
$$
\left[\left(\frac{a-b}{x+x}\right)^{-3}\right]^{2}\left[\left(\frac{x-y}{b-a}\right)^{-2}\right]^{3}\left[\left(\frac{x-y}{x+y}\right)^{-2}\right]^{-3}=?
$$