Change vectors condition and still make the equation true For a equation with random vector, this is true: $(v\times w)\times z=c_{1}v+c_{2}v+c_{3}(v\times w)$. The symbol $ \times $ is the cross product, and $v, w, z \in \mathbb{R^{3}}$, $c_{1}, c_{2}, c_{3} \in \mathbb{R}$. Then suppose that $v$ is changed to $v=0$. But every vector not $v$ is not $0$. Then how do I make sure that the equation is still true?
Similarly, if I suddenly change $v=0$ and $w=0$, $v=mw$, $m \not =0$(the vectors is a multiple,) then how do I still make the equation true?
 A: By my understanding, you are trying to prove that the following statement:

For any $  v ,w,z \in \mathbb{R}^3 $, there are real scalars $ c_1, c_2, c_3$ such that
$$ \tag{*} \label{eq} (v \times w) \times z = c_1 v + c_2 w + c_3 (v \times w).  $$

To prove this, you may separate into several cases first, as you suggested.
Case 1. $v$ and $w$ are linearly dependent.
In this case, one may write $ v = kw$ for some real scalar $k$. Notice that $ v \times w $ = 0 in this case. So the left hand side of \eqref{eq} is zero. One can then pick $c_1 = c_2 = c_3 = 0 $.
Case 2. $v$ and $w$ are linearly independent.
In this case, $v \times w \neq 0$. In addition, $ v \times w $ is orthogonal to both $ v $ and $w$. In particular, the set of vectors $ \lbrace v, w, v \times w \rbrace$ is linearly independent. Since the dimension of $\mathbb{R}^3 $ is 3, the set $ \lbrace v, w, v \times w \rbrace$ forms a basis of $\mathbb{R}^3$. It follows that $ \lbrace v, w, v \times w \rbrace $ spans $\mathbb{R}^3$. \eqref{eq} follows by the definition of a spanning set.
