Question

Show that
$\left|\begin{array}{cccccc}a& b& c& d& e& f\\ f& a& b& c& d& e\\ e& f& a& b& c& d\\ d& e& f& a& b& c\\ c& d& e& f& a& b\\ b& c& d& e& f& a\end{array}\right|=\left|\begin{array}{ccc}A& B& C\\ C& A& B\\ B& C& A\end{array}\right|$,

where

$A=a^2-d^2+2ce-2bf$,
$B=e^2-b^2+2ac-2df$,
$C=c^2-f^2+2ae-2bd$.

Answer 1

Adding together all the columns we see that the determinant is divisible by $a+b+c+d+e+f$

Multiplying the columns by $1,-1,1,-1,1,-1$ respectively and adding the results we see that

$a-b+c-d+e-f$ is a factor of the determinant

Multiplying the columns by $1,w,w^2,1,w,w^2$ respectively and adding the it follows that

$a+wb+w^{2}e+d+ew+f{w}^2$

Similarly we may show that

$a+w^{2}b+wc+d+w^{2}e+wf$

$a-wb+w^{2}c-d+we-w^{2}f$

$a-w^{2}b+ec-d+w^{2}e-wf;$

are factors of the determinant.

Hence the determinant is the product of these six factors and some constant which is unity

Taking these factors in pairs, it follows that the determinant is the product of the three expressions.

${(a+c+e)}^2-{(b+d+f)}^2$

${(a+w^{2}c+we)}^2-{(d+w^{2}f+wb)}^2;$

${(a+wc+w^{2}e)}^2-{(d+wf+w^{2}b)}^2$

The last of these factors

$=a^2-d^2+2ce-2bf+w(e^2-b^2+2ac-2df)+w^2(c^2-f^2+2ae-2bd)$

$=A+wB+w^{2}C$

Similarly the second factor

$A+w^{2}B+wC$

and the first factor$=A+B+C$

Hence the determinant

$=(A+B+C)(A+wB+w^{2}C)(A+w^{2}B+wC)$

$=A^3+B^3+C^3+-ABC$

$=\left|\begin{array}{ccc}A& B& C\\ C& A& B\\ B& C& A\end{array}\right|$