In my textbook, the author claims that the following can be proved by chaining vector triple product and scalar triple product
$$\text{i.) }(A \times B) \cdot (C \times D) = (A \cdot C)(B \cdot D) - (A \cdot D)(B \cdot C)$$
$$\text{ii.) }A \times [B \times (C \times D)] = B[A \cdot (C \times D)] - (A \cdot B)(C \times D)$$
I can prove both by writing out the component form but cannot derive the identity by chaining the rules
for example
$$A \times [B \times (C \times D)] = A \times (C(B \cdot D) - D(B \cdot C)) $$
$$= A \times (C(B \cdot D)) - A \times (D(B \cdot C))$$
after that triple product rule is non applicable, am I missing something?
(i) Applying the symmetry $$X \cdot (Y \times Z) = Y \cdot (Z \times X)$$ to $X = A \times B, Y = C, Z = D$ gives $$(A \times B) \cdot (C \times D) = (D \times (A \times B)) \cdot C.$$ Now, applying the triple vector product identity $$D \times (A \times B) = A (D \cdot B) - B (D \cdot A)$$ gives that the right-hand side is $$[A (D \cdot B) - B (D \cdot A)] \cdot C] = (A \cdot C)(B \cdot D) - (A \cdot D)(B \cdot A)$$ as desired.
(ii) Applying the triple cross product identity to $A$, $B$, and $C \times D$ gives that the left-hand side is $$A \times (B \times (C \times D))) = B(A \cdot (C \times D)) - (C \times D)(A \cdot B).$$