Let $\sigma$ be the permutation $\sigma=(1234)(35)(2653)^2(56)$. I've got to say whether $\sigma$ is a product of 3-cycles or not.
My attempt was to simplify $(2653)^2=(36)(25)=(25)(36)$. Is this correct? At the end I'd get $\sigma=(63)(5)(241)$, which shows that it can't be written as a product of 3-cycles. I hope that's right. Thanks for helping.
Unfortunately, just because something can be written in simplest form without $3$-cycles doesn't mean that it's not a composition of $3$-cycles; $(12)(34)$ is in fact a composition of two $3$-cycles (which I leave as an exercise; in fact, all even permutations can be formed from composing $3$-cycles).
A better proof here is to use permutation parity; $\sigma$ is an odd permutation (by summing parities of the permutations that make it), but $3$-cycles are all even permutations.