There are an infinite number of cube-free binary words.  Let $w$ be one such word (such as the Thue-Morse word).  Let $w-j$ denote word obtained by deleting the first $j$ letters from $w$.  Clearly, for all $j \in \mathbb{N}$, $w-j$ is also cube-free.  Also, I claim that $w-j \neq w-k$ for any $j < k$, from which the result follows. To see this, let $w=abc$, where $a$ has length $j$ and $b$ has length $k-j$.  Then $abc=ababc$ by hypothesis.  Iterating again, we see that $w$ starts with $ababab$, a contradiction.