I watched a video about vector triple product expansion. They were proving it. At the end of the video they obtained the following formula which you can access it by the link.
According to formula, $$\vec a×(\vec b×\vec c)=\vec b(\vec a·\vec c)-\vec c(\vec a·\vec b)$$ Also this is always equals $0$ because $\vec b(\vec a·\vec c)$ and $\vec c(\vec a·\vec b)$ are identical. Am I wrong or is video wrong?
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In short, $\vec{b}(\vec{a} \cdot \vec{c})$ is not the same as $\vec{c}(\vec{a} \cdot \vec{b})$, because $\vec{a} \cdot \vec{c}$ and $\vec{a} \cdot \vec{b}$ are scalers, not vectors.
If $a$, $b$ and $c$ are scalers, then
$b(a \cdot c) = b(ac) = bac = cab = c(ab) = c(a \cdot b)$
But, the same is not true for vectors. If we did the same with vectors (in this case $\vec{a}$, $\vec{b}$ and $\vec{c}$), then $\vec{b}(\vec{a} \cdot \vec{c})$ would be $\vec{b}$ multiplied by the scaler $\vec{a} \cdot \vec{c}$ and $\vec{c}(\vec{a} \cdot \vec{b})$ would be the same as $\vec{c}$ multiplied by $\vec{a} \cdot \vec{b}$. (If you're not sure about why these are scalers, you can look at "Dot products and duality | Chapter 9, Essence of linear algebra", a good video by 3Blue1Brown or just Google the definition of dot products.)
Therefore, it is not true that $\vec{b}(\vec{a} \cdot \vec{c}) = \vec{c}(\vec{a} \cdot \vec{b})$ for all vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$, because $\vec{b}$ isn't the same as $\vec{c}$ multiplied by a scaler for all $\vec{b}$ and $\vec{c}$. (Just to be clear, that's not to say anything about whether $\vec{b}(\vec{a} \cdot \vec{c}) = \vec{c}(\vec{a} \cdot \vec{b})$ holds for some values of $\vec{b}$ and $\vec{c}$, if $\vec{b}$ was $\vec{c}$ multiplied by a scaler.)
Your statement that "$\vec{b}(\vec{a}\cdot\vec{c})$ and $\vec{c}(\vec{a}\cdot\vec{b})$ are identical" is simply WRONG! The first is a vector in the same direction as $\vec{b}$ and the second is a vector in the same direction as $\vec{a}$.