Prove the BAC-CAB rule by writing out both sides in component form.

To proveBAC-CAB rule, $A\times B\times C=B(A.C)-C(A.B)$, firstly evaluate its left side,and right side saperately, then the results of both sides mst be compared.

Let the vectors are defined as $A=A_{x}i+A_{y}j+A_{z}k,B=B_{x}i+B_{y}j+B_{z}k$, and$C=C_{x}i+C_{y}j+C_{z}k$

$$\begin{gathered} B\times C=\left|\begin{array}{ccc}i& j& k\\ B_x& B_y& B_z\\ C_x& C_y& C_z\end{array}\right| \\ =\left(B_y{C}_z-B_z{C}_y\right)\text{i -}\left(C_z-B_z-C_x\right)\text{j +}\left(B_x{C}_y-B_y{C}_x\right)k \end{gathered}$$

Now use the above result to evaluate A x B x C as,

$\begin{array}{rcl}A\times B\times C& =& \left|\begin{array}{ccc}i& j& k\\ A_x& A_y& A_z\\ B_y{C}_z-B_z{C}_y& B_x{C}_z-B_z{C}_x& B_x{C}_y-B_y{C}_x\end{array}\right|\\ & =& \left(A_y{B}_x{C}_y-A_y{B}_y{C}_x-A_z{B}_z{C}_x\text{+}{A}_z{B}_x{C}_z\right)i\\ & & -\left(-A_x{B}_x{C}_y-A_x{B}_y{C}_x-A_z{B}_y{C}_z+A_z{B}_z{C}_y\right)j\\ & & +\left(A_x{B}_z{C}_x-A_x{B}_x{C}_z-A_y{B}_y{C}_z+A_y{B}_z{C}_y\right)k\\ & & \end{array}$

Substitute $A_{x}i+A_{y}j+A_{z}k$for A, $B_{x}i+B_{y}j+B_{z}k$for B, and $C_{x}i+C_{y}j+C_{z}k$for C into$B\left(A·C\right)-C\left(A·B\right)$

$$\begin{gathered} B\left(A·C\right)-C\left(A·B\right)=\left(B_x{A}_y{C}_y-C_x{A}_y{B}_y-C_x{A}_z{B}_z+B_x{A}_z{C}_z\right)\text{i +} \\ \left(-C_y{A}_x{B}_x+B_y{A}_x{C}_x+B_y{A}_z{C}_z-C_y{A}_z{B}_z\right)\text{j}+ \\ \left(B_z{A}_x{C}_x-C_z{A}_y{B}_y-C_z{A}_y{B}_y+B_z{A}_y{C}_y\right)k \end{gathered}$$ …… (2)

From equation (1) and (2) it can be concluded that the result of $B\left(A·C\right)-C\left(A·B\right)$and $A\times B\times C$ is same. Thus $A\times B\times C=B\left(A·C\right)-C\left(A·B\right)$