Let A, B, C and D be eightdimensional vectors. The problem here is to solve A × B × C = D for C, when A, B and D are furnished.
First define these 28 real numbers:
h_{10} = a_{3}b_{2} − a_{2}b_{3} + a_{5}b_{4} − a_{4}b_{5} + a_{7}b_{6} − a_{6}b_{7}
h_{20} = a_{1}b_{3} − a_{3}b_{1} + a_{5}b_{7} − a_{7}b_{5} + a_{6}b_{4} − a_{4}b_{6}
h_{30} = a_{2}b_{1} − a_{1}b_{2} + a_{4}b_{7} − a_{7}b_{4} + a_{5}b_{6} − a_{6}b_{5}
h_{40} = a_{1}b_{5} − a_{5}b_{1} + a_{2}b_{6} − a_{6}b_{2} + a_{7}b_{3} − a_{3}b_{7}
h_{50} = a_{4}b_{1} − a_{1}b_{4} + a_{6}b_{3} − a_{3}b_{6} + a_{7}b_{2} − a_{2}b_{7}
h_{60} = a_{1}b_{7} − a_{7}b_{1} + a_{3}b_{5} − a_{5}b_{3} + a_{4}b_{2} − a_{2}b_{4}
h_{70} = a_{2}b_{5} − a_{5}b_{2} + a_{3}b_{4} − a_{4}b_{3} + a_{6}b_{1} − a_{1}b_{6}
h_{21} = a_{3}b_{0} − a_{0}b_{3} + a_{6}b_{5} − a_{5}b_{6} + a_{7}b_{4} − a_{4}b_{7}
h_{31} = a_{0}b_{2} − a_{2}b_{0} + a_{5}b_{7} − a_{7}b_{5} + a_{6}b_{4} − a_{4}b_{6}
h_{41} = a_{2}b_{7} − a_{7}b_{2} + a_{3}b_{6} − a_{6}b_{3} + a_{5}b_{0} − a_{0}b_{5}
h_{51} = a_{0}b_{4} − a_{4}b_{0} + a_{2}b_{6} − a_{6}b_{2} + a_{7}b_{3} − a_{3}b_{7}
h_{61} = a_{4}b_{3} − a_{3}b_{4} + a_{5}b_{2} − a_{2}b_{5} + a_{7}b_{0} − a_{0}b_{7}
h_{71} = a_{0}b_{6} − a_{6}b_{0} + a_{3}b_{5} − a_{5}b_{3} + a_{4}b_{2} − a_{2}b_{4}
h_{32} = a_{1}b_{0} − a_{0}b_{1} + a_{4}b_{5} − a_{5}b_{4} + a_{6}b_{7} − a_{7}b_{6}
h_{42} = a_{5}b_{3} − a_{3}b_{5} + a_{6}b_{0} − a_{0}b_{6} + a_{7}b_{1} − a_{1}b_{7}
h_{52} = a_{0}b_{7} − a_{7}b_{0} + a_{3}b_{4} − a_{4}b_{3} + a_{6}b_{1} − a_{1}b_{6}
h_{62} = a_{0}b_{4} − a_{4}b_{0} + a_{1}b_{5} − a_{5}b_{1} + a_{7}b_{3} − a_{3}b_{7}
h_{72} = a_{1}b_{4} − a_{4}b_{1} + a_{3}b_{6} − a_{6}b_{3} + a_{5}b_{0} − a_{0}b_{5}
h_{43} = a_{0}b_{7} − a_{7}b_{0} + a_{2}b_{5} − a_{5}b_{2} + a_{6}b_{1} − a_{1}b_{6}
h_{53} = a_{0}b_{6} − a_{6}b_{0} + a_{1}b_{7} − a_{7}b_{1} + a_{4}b_{2} − a_{2}b_{4}
h_{63} = a_{1}b_{4} − a_{4}b_{1} + a_{2}b_{7} − a_{7}b_{2} + a_{5}b_{0} − a_{0}b_{5}
h_{73} = a_{4}b_{0} − a_{0}b_{4} + a_{5}b_{1} − a_{1}b_{5} + a_{6}b_{2} − a_{2}b_{6}
h_{54} = a_{1}b_{0} − a_{0}b_{1} + a_{2}b_{3} − a_{3}b_{2} + a_{6}b_{7} − a_{7}b_{6}
h_{64} = a_{2}b_{0} − a_{0}b_{2} + a_{3}b_{1} − a_{1}b_{3} + a_{7}b_{5} − a_{5}b_{7}
h_{74} = a_{0}b_{3} − a_{3}b_{0} + a_{2}b_{1} − a_{1}b_{2} + a_{5}b_{6} − a_{6}b_{5}
h_{65} = a_{0}b_{3} − a_{3}b_{0} + a_{2}b_{1} − a_{1}b_{2} + a_{4}b_{7} − a_{7}b_{4}
h_{75} = a_{0}b_{2} − a_{2}b_{0} + a_{1}b_{3} − a_{3}b_{1} + a_{6}b_{4} − a_{4}b_{6}
h_{76} = a_{1}b_{0} − a_{0}b_{1} + a_{2}b_{3} − a_{3}b_{2} + a_{4}b_{5} − a_{5}b_{4}
The pattern is that a_{i} b_{j} contributes to h_{kl} when U_{i} × U_{j} × U_{k} = U_{l} accoring to the baseline cross product. Should h_{kl} not be listed, it equals − h_{lk}. Meanwhile, h_{kk} had it been needed would have equalled zero.
Now form the skewsymmetric matrix H:
H = 

Then, with the double dot indicating matrix multiplication, seek values of C that satisfy H ·· C = D. This problem is equivalent to solving a system of eight linear equations in eight variables. Not yet established is whether H is invertible, therefore there may be zero, one or many solutions.
Here is an example where three vectors are given:
A = [  +1,  0,  +1,  −1,  0,  +1,  0,  −1 ] 
B = [  −1,  0,  0,  +1,  −1,  +1,  −1,  0 ] 
D = [  +3,  0,  −7,  −4,  −4,  +2,  −1,  +2 ] 
and the goal is to find C in A × B × C = D.
The coefficients in the following system correspond to the values in the H matrix above, and the right member of each equation comes from D:
0c_{0}  − 1c_{1}  + 1c_{2}  − 2c_{3}  − 2c_{4}  − 1c_{5}  − 1c_{6}  + 2c_{7} =  +3 
1c_{0}  + 0c_{1}  + 2c_{2}  + 2c_{3}  − 1c_{4}  − 3c_{5}  − 1c_{6}  − 2c_{7} =  0 
− 1c_{0}  − 2c_{1}  + 0c_{2}  + 0c_{3}  + 3c_{4}  + 0c_{5}  − 2c_{6}  − 1c_{7} =  −7 
2c_{0}  − 2c_{1}  + 0c_{2}  + 0c_{3}  + 0c_{4}  + 0c_{5}  − 2c_{6}  + 2c_{7} =  −4 
2c_{0}  + 1c_{1}  − 3c_{2}  + 0c_{3}  + 0c_{4}  + 0c_{5}  − 2c_{6}  − 1c_{7} =  −4 
1c_{0}  + 3c_{1}  + 0c_{2}  + 0c_{3}  + 0c_{4}  + 0c_{5}  − 1c_{6}  + 1c_{7} =  +2 
1c_{0}  + 1c_{1}  + 2c_{2}  + 2c_{3}  + 2c_{4}  + 1c_{5}  + 0c_{6}  + 2c_{7} =  −1 
− 2c_{0}  + 2c_{1}  + 1c_{2}  − 2c_{3}  + 1c_{4}  − 1c_{5}  − 2c_{6}  + 0c_{7} =  +2 
With application of Gaussian elimination, two equations drop out, leaving an underdetermined system:
c_{3} + c_{6} − c_{7} =  0 
c_{5} + c_{6} + c_{7} =  +1 
c_{4} − c_{6} =  −2 
c_{2} + c_{7} =  +1 
c_{0} − c_{6} + c_{7} =  −1 
c_{1} =  +1 
Among the many parameterizations of C is:
C = [ c_{6} − c_{7} − 1, 1, 1 − c_{7}, c_{7} − c_{6}, c_{6} − 2, 1 − c_{6} − c_{7}, c_{6}, c_{7} ]
which, being linear and having two independent variables, namely c_{6} and c_{7}, denotes a plane. A particular solution is C = [ 0, +1, +1, −1, −1, 0, +1, 0 ].
If A × B × C = 0, but A and B are nonzero, then C must be a linear combination of A and B, also governed by two independent variables.
We prepared a computer program to evaluate the determinant of matrix H by symbolic manipulation (as opposed to numerical techniques). It turns out that for all complex values of a_{n} and b_{n}, the determinant vashishes.
Although evaluating the determinant of an 8by8 matrix most generally requires calculating 8! = 40320 products of 8 factors each, we bypassed any product that would have incorporated a zero element on the diagonal, reducing the number of products to 14833 (see derangement). Beyond that, we did not take advantage of the skew symmetry of the matrix, preferring to keep the algorithm simple to facilitate checking.
An example of a product is:
h_{30}h_{51}h_{12}h_{23}h_{04}h_{45}h_{76}h_{67}
which is a polynomial in 214207 terms. (Recall that h_{67} = −h_{76}) An example of a term in that product is:
−2a_{1}(a_{2})^{3}(a_{5})^{2}a_{6}a_{7}b_{1}b_{2}(b_{3})^{2}(b_{6})^{3}b_{7}
In all terms of all products, the sum of exponents of all the a_{n} is eight, and no exponent is greater than six. The same applies to the b_{n}.
A standard result is that the rank of a skewsymmetric matrix must be an even number.