$\mathbf{A}\times \left(\mathbf{B}\+\mathbf{C}\right)$

$\= \|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ b_1\+c_1& b_2\+c_2& b_3\+c_3\end{array}\|$ = $\left[\begin{array}{c}{a}_2\left(b_3\+c_3\right)-a_3\left(b_2\+c_2\right)\\ a_3\left(b_1\+c_1\right)-a_1\left(b_3\+c_3\right)\\ a_1\left(b_2\+c_2\right)-a_2\left(b_1\+c_1\right)\end{array}\right]$

$\mathbf{A}\times \mathbf{B}\+\mathbf{A}\times \mathbf{C}$

$\= \left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\|\+\right|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ c_1& c_2& c_3\end{array}\|$

$\=\left[\begin{array}{c}{a}_2{b}_3-a_3{b}_2\\ a_3{b}_1-a_1{b}_3\\ a_1{b}_2-a_2{b}_1\end{array}\right]\+\left[\begin{array}{c}{a}_2{c}_3-a_3{c}_2\\ a_3{c}_1-a_1{c}_3\\ a_1{c}_2-a_2{c}_1\end{array}\right]$ = $\left[\begin{array}{c}{a}_2\left(b_3\+c_3\right)-a_3\left(b_2\+c_2\right)\\ a_3\left(b_1\+c_1\right)-a_1\left(b_3\+c_3\right)\\ a_1\left(b_2\+c_2\right)-a_2\left(b_1\+c_1\right)\end{array}\right]$

$\left(\mathbf{A}\+\mathbf{B}\right)\times \mathbf{C}$

$\= \|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1\+b_1& a_2\+b_2& a_3\+b_3\\ c_1& c_2& c_3\end{array}\|\=\left[\begin{array}{c}\left(a_2\+b_2\right)c_3-\left(a_3\+b_3\right)c_2\\ \left(a_3\+b_3\right)c_1-\left(a_1\+b_1\right)c_3\\ \left(a_1\+b_1\right)c_2-\left(a_2\+b_2\right)c_1\end{array}\right]$

$\mathbf{A}\times \mathbf{C}\+\mathbf{B}\times \mathbf{C}$

$\=$$\|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ a_1& a_2& a_3\\ c_1& c_2& c_3\end{array}\|\+\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$ 

$\=\left[\begin{array}{c}{a}_2{c}_3-a_3{c}_2\\ a_3{c}_1-a_1{c}_3\\ a_1{c}_2-a_2{c}_1\end{array}\right]\+\left[\begin{array}{c}{b}_2{c}_3-b_3{c}_2\\ b_3{c}_1-b_1{c}_3\\ b_1{c}_2-b_2{c}_1\end{array}\right]$ = $\left[\begin{array}{c}\left(a_2\+b_2\right)c_3-\left(a_3\+b_3\right)c_2\\ \left(a_3\+b_3\right)c_1-\left(a_1\+b_1\right)c_3\\ \left(a_1\+b_1\right)c_2-\left(a_2\+b_2\right)c_1\end{array}\right]$

Define the vectors A, B, and C 

$⟨a_1\,a_2\,a_3⟩$$\stackrel{\text{assign to a name}}{\to }$$A$

$⟨b_1\,b_2\,b_3⟩$$\stackrel{\text{assign to a name}}{\to }$$B$

$⟨c_1\,c_2\,c_3⟩$$\stackrel{\text{assign to a name}}{\to }$$C$

Show $\mathbf{A}\times \left(\mathbf{B}\+\mathbf{C}\right)-\left(\mathbf{A}\times \mathbf{B}\+\mathbf{A}\times \mathbf{C}\right)\=\mathbf{0}$ 

$\mathbf{A}\times \left(\mathbf{B}\+\mathbf{C}\right)-\left(\mathbf{A}\times \mathbf{B}\+\mathbf{A}\times \mathbf{C}\right)$

$\stackrel{\text{simplify}}{\=}$

Show $\left(\mathbf{A}\+\mathbf{B}\right)\times \mathbf{C}- \left(\mathbf{A}\times \mathbf{C}\+\mathbf{B}\times \mathbf{C}\right)\=\mathbf{0}$ 

$\left(\mathbf{A}\+\mathbf{B}\right)\times \mathbf{C}-\left(\mathbf{A}\times \mathbf{C}\+\mathbf{B}\times \mathbf{C}\right)$

$\stackrel{\text{simplify}}{\=}$

Initialize

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right)\:$

$\mathbf{A}\,\mathbf{B}\,\mathbf{C}≔⟨a_1\,a_2\,a_3⟩\,⟨b_1\,b_2\,b_3⟩\,⟨c_1\,c_2\,c_3⟩\:$

Obtain and test the two sides of the first distributive law

Left-hand Side

Right-hand Side

${\mathrm{LHS}}_1≔\mathrm{CrossProduct}\left(\mathbf{A}\,\mathbf{B}\+\mathbf{C}\right)$

${\mathrm{RHS}}_1≔\mathrm{CrossProduct}\left(\mathbf{A}\,\mathbf{B}\right)\+\mathrm{CrossProduct}\left(\mathbf{A}\,\mathbf{C}\right)$

$\mathrm{simplify}\left({\mathrm{LHS}}_1-{\mathrm{RHS}}_1\right)$ = $$

Obtain and test the two sides of the second distributive law

Left-hand Side

Right-hand Side

${\mathrm{LHS}}_2≔\mathrm{CrossProduct}\left(\mathbf{A}\+\mathbf{B}\,\mathbf{C}\right)$

${\mathrm{RHS}}_2≔\mathrm{CrossProduct}\left(\mathbf{A}\,\mathbf{C}\right)\+\mathrm{CrossProduct}\left(\mathbf{B}\,\mathbf{C}\right)$

$\mathrm{simplify}\left({\mathrm{LHS}}_2-{\mathrm{RHS}}_2\right)$ = $$