I am trying to prove that $A\times(B+C)=A \times B + A \times C.$
I have managed to prove this by finding the LHS $$\begin{pmatrix} a_{1}\\a_{2}\\a_{3} \end{pmatrix} × \begin{pmatrix} b_{1}+c_{1} \\b_{2}+c_{2}\\b_{3}+c_{3} \end{pmatrix}$$ then finding the RHS $$\begin{pmatrix} a_{1}\\a_{2}\\a_{3} \end{pmatrix} × \begin{pmatrix} b_{1} \\b_{2}\\b_{3} \end{pmatrix} + \begin{pmatrix} a_{1} \\a_{2}\\a_{3} \end{pmatrix}× \begin{pmatrix} c_{1} \\c_{2}\\c_{3} \end{pmatrix}$$and equating LHS = RHS.
However I need to prove the equation directly from the LHS rather than finding the RHS and equating (possibly using separate matrix properties). I.E. I need to continue proving from \begin{pmatrix} a_{2}b_{3}+a_{2}c_{3} - a_{3}b_{2} - a_{3}c_{2} \\a_{3}b_{1} + a_{3}c_{1} - a_{1}b_{3}-a_{1}c_{3}\\a_{1}b_{2}+a_{1}c_{2} -a_{2}b_{1}-a_{2}c_{1} \end{pmatrix}
Notice that: $$\begin{pmatrix} a_{2}b_{3}+a_{2}c_{3} - a_{3}b_{2} - a_{3}c_{2} \\a_{3}b_{1} + a_{3}c_{1} - a_{1}b_{3}-a_{1}c_{3}\\a_{1}b_{2}+a_{1}c_{2} -a_{2}b_{1}-a_{2}c_{1} \end{pmatrix}=\begin{pmatrix} a_{2}b_{3} - a_{3}b_{2} \\a_{3}b_{1} - a_{1}b_{3}\\a_{1}b_{2} -a_{2}b_{1} \end{pmatrix}+\begin{pmatrix}a_2c_3-a_3c_2\\a_3c_1-a_1c_3\\a_1c_2-a_2c_1\end{pmatrix}$$
And this is precisely the RHS.