Tensor notation (practicing)
There is no mistake, your solution is correct. Look for a variable substitution:
$$A = C' \qquad B = D' \qquad C = A' \qquad D = B'$$
Your result show that: $$ (C'\times D')\times (A' \times B') = \det(C',D',B')A' - \det(C',D',A')B' $$
Let's just forget lexicografic fomallity and rewrite this as
$$ (C\times D)\times(A \times B) = \det(B,C,D)A - \det(A,C,D)B $$
which is the same you wanted first, since $$(C\times D)\times(A \times B) = -(A\times B)\times(C\times D) $$