Proving that $(u+v)×w=u×w+v×w$ Let's $$(\overrightarrow{u}+\overrightarrow{v})\times\overrightarrow{w}=\overrightarrow{u}\times\overrightarrow{w}+\overrightarrow{v}\times\overrightarrow{w}$$ How to prove it?

Update: The problem is that I don't know how to do it because I don't know the number of dimensions. Is there a general formula?
This property should do it. Let $u,v,w\in V$ (where we take $V$ a presumably three dimensional inner product space). Let $\langle\cdot,\cdot\rangle$ be the inner product on $V$. Then $$\langle u,v\times w\rangle =\det[u,v,w]_{\mathcal B}$$

Where $\det[\,]_{\mathcal B}$ denotes the determinant of the matrix formed by $u,v,w$ in an orthonormal basis $\mathcal B = \{e_1,e_2,e_3\}$. 

In this light let, $\gamma_1$ be the first coordinate of $\gamma = (a+b)\times w$. Then, $$\gamma_1 = \langle e_1,(a+b)\times w\rangle = \det[e_1,a+b,w] = \det[e_1,a,w] + \det[e_1,b,w] =  \\ =  \langle e_1,a\times w\rangle + \langle e_1, b\times w\rangle$$

Idem for $\gamma_2,\gamma_3$. Can you take it from here?

P.S. Take $\mathbb R^3$ with the usual scalar product if the notation here clouds things up.
The cross product is defined on two vectors of three dimensions as: $$\vec a \times \vec b =\begin{bmatrix} a_1 \\a_2\\a_3\end{bmatrix}\times \begin{bmatrix} b_1 \\b_2\\b_3 \end{bmatrix} = \begin{bmatrix} a_2b_3-b_3a_2 \\a_3b_1-a_1b_3\\a_1b_2-b_2a_1 \end{bmatrix}$$

The sum of two three-dimensional vectors is:
$$\vec c + \vec d =\begin{bmatrix} c_1 \\c_2\\c_3\end{bmatrix}+ \begin{bmatrix} d_1 \\d_2\\d_3 \end{bmatrix} = \begin{bmatrix} c_1+d_1 \\c_2+d_2\\c_3+d_3 \end{bmatrix}$$

Use this to show that: $$(\vec u+\vec v)\times \vec w = (\vec u\times \vec w)+(\vec v\times \vec w)$$