Actually, it's "fully symmetric" which is *different* than cyclic symmetry. To see the difference, consider:
$(a - b)(b - c)(c - a)$
If we send $a \to b \to c \to a$, we get:
$(b - c)(c - a)(a - b)$, which is the same expression (even though the factors are in a different order).
If we just switch $a$ and $b$, we get
$(b - a)(a - c)(c - b) = -(a - b)(b - c)(c - a)$, which is *not* the original expression, but its negative.
That is-if we can swap any pair and get the original expression, it is fully symmetric (permutations are generated by pair-swaps), full symmetry implies cyclic symmetry, but the reverse is not so.
(Note that $(a - b)(b - c)(c - a)$ also remains unchanged under the transformation $a \to c \to b \to a$ which gives:
$(c - a)(a - b)(b - c)$).
The transformation (substitution) $a \to b \to c \to a$ is called a 3-cycle, as it changes 3 things to something else, and repeating it three times "completes the cycle" and leaves you where you were.