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. Towards a contradiction, suppose that $w-j=w-k$ for some $j < k$. Let $w=abc$, where $a$ has length $j$ and $b$ has length $k-j$. Then $bc=c$ by hypothesis. Iterating, we see that $c$ starts with $bbb$, contradicting that $w$ is cube-free.

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 the 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. Towards a contradiction, suppose that $w-j=w-k$ for some $j < k$. Let $w=abc$, where $a$ has length $j$ and $b$ has length $k-j$. Then $bc=c$ by hypothesis. Iterating, we see that $c$ starts with $bbb$, contradicting that $w$ is cube-free.

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.

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. Towards a contradiction, suppose that $w-j=w-k$ for some $j < k$. Let $w=abc$, where $a$ has length $j$ and $b$ has length $k-j$. Then $bc=c$ by hypothesis. Iterating, we see that $c$ starts with $bbb$, contradicting that $w$ is cube-free.