We know that
\begin{equation} \vec{A}\circ(\vec{B}\times\vec{C})=\vec{B}\circ(\vec{C}\times\vec{A})=\vec{C}\circ(\vec{A}\times\vec{B}) \qquad (1). \end{equation}
In some reference I found that
\begin{equation} \vec{A}\circ(\vec{B}\times\vec{C})=\vec{C}\circ(\vec{A}\times\vec{B})-\vec{B}\circ(\vec{A}\times\vec{C})\qquad (2). \end{equation}
Based on (1), we can substitute, in (2), $\vec{B}\circ(\vec{C}\times\vec{A})$ for $\vec{C}\circ(\vec{A}\times\vec{B})$, then $\vec{A}\circ(\vec{B}\times\vec{C})=\vec{B}\circ(\vec{C}\times\vec{A})-\vec{B}\circ(\vec{A}\times\vec{C})$. And since $\vec{B}\circ(\vec{A}\times\vec{C})=-\vec{B}\circ(\vec{C}\times\vec{A})$, then \begin{equation} \vec{A}\circ(\vec{B}\times\vec{C})=2\left(\vec{B}\circ(\vec{C}\times\vec{A})\right). \end{equation}
We also get \begin{equation} \vec{A}\circ(\vec{B}\times\vec{C})=2\left(\vec{C}\circ(\vec{A}\times\vec{B})\right) \end{equation} by substituting, in (2), $-\vec{C}\circ(\vec{A}\times\vec{B})$ for $\vec{B}\circ(\vec{A}\times\vec{C})$.
It's clear that the obtained results doesn't match the equalities in (1), unless (2) becomes \begin{equation} \vec{A}\circ(\vec{B}\times\vec{C})=\frac{1}{2}\left(\vec{C}\circ(\vec{A}\times\vec{B})-\vec{B}\circ(\vec{A}\times\vec{C})\right) \qquad (3). \end{equation} So what's wrong ?
Note that, in many books of vector calculus one can find the equality \begin{equation} \vec{\nabla}\circ(\vec{U}\times\vec{V})=\vec{V}\circ(\vec{\nabla}\times\vec{U})-\vec{U}\circ(\vec{\nabla}\times\vec{V})\qquad (4) \end{equation} which means that (2) is true without $\frac{1}{2}$, just by substituting $\vec{\nabla}$ for $\vec{A}$, $\vec{U}$ for $\vec{B}$, and $\vec{V}$ for $\vec{C}$.
Presumably there is an equals sign missing in your reference and it's meant to read
$$\vec{A}\cdot(\vec{B}\times\vec{C})=\vec{C}\cdot(\vec{A}\times\vec{B})=-\vec{B}\cdot(\vec{A}\times\vec{C})$$
UPDATE to address questions raised in the comments: for a counterexample, note that
$$(1,0,0) \cdot \bigg[(0,1,0) \times (0,0,1)\bigg] = (1,0,0) \cdot (1,0,0) = 1$$
whereas
\begin{align*} (0,0,1) &\cdot \bigg[(1,0,0) \times (0,1,0)\bigg] - (0,1,0) \cdot \bigg[(1,0,0) \times (0,0,1)\bigg] \\ &= (0,0,1) \cdot (0,0,1) - (0,1,0) \cdot (0, -1, 0) \\ &= 1 - (-1) = 2 \end{align*}
To address the apparent similarity to the product rule for divergence, $$\vec{\nabla}\cdot(\vec{U}\times\vec{V})=\vec{V}\cdot(\vec{\nabla}\times\vec{U})-\vec{U}\cdot(\vec{\nabla}\times\vec{V}),$$ let's look at what happens in the one-dimensional case. Suppose that $a, b, c$ are real numbers. Then
$$a b c = c a b = b a c$$
so in particular we have
$$a b c = \frac{1}{2} (b a c + c a b).$$
On the other hand, suppose that $u$ and $v$ are functions of $x$. Then by the product rule,
$$\frac{d}{dx} \; u \; v = u \; \frac{d}{dx} \; v + v \; \frac{d}{dx} \; u$$
The preceding two equations look very similar except for the factor of 1/2, but clearly they do not actually have much to do with one another. In particular, the chain rule from Calculus 1 does not come from taking the first equation, substituting $a = d/dx$, and somehow dealing with the factor of 2. The apparent similarity is essentially a coincidence.
So we should ask ourselves: what's the difference here, at a fundamental level? Well, multiplication of real numbers (or functions) is commutative, but "multiplication" by the differential operator $\frac{d}{dx}$ is not; $a b c$ is the same thing as $b a c$, but $$\frac{d}{dx} \; u \; v$$ is not the same thing as $$u \; \frac{d}{dx} \; v$$
The three-dimensional case you're looking at here is entirely analogous to the one-dimensional case I've described above, with "commutative" replaced by "anti-commutative." In fact, if you go ahead and expand your (1) and (3) in coordinates, you'll see it essentially reduces to the one-dimensional case.
As to why certain identities among vectors hold when replacing a vector with a $\vec{\nabla}$: not every identity makes use of the commutativity of multiplication in a fundamental way.