Problem on Determinant. Q.
$$\text{If } \Delta = \left|\begin{array}{ccc} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|,$$ $$\text{ then the value of }$$ $$ \left|\begin{array}{ccc} a^2 - bc & b^2 - ca & c^2 - ab \\ c^2 - ab & a^2 - bc & b^2 - ca \\ b^2 - ca & c^2 - ab & a^2 - bc \end{array}\right| \text{ is:}$$
Express the Answer in terms of $ \Delta $.
My Attempt - I tried Rearranging the Second determinant (by adding or subtracting particular rows or Columns) such that after this, It'll be easy for me write the second determinant as the product of two or more other Determinants that might be same as $\Delta$. But I Found that rearranging the Second Det. Wasn't a good choice as it seems to never simplify itself!! So, I tried Expressing the Second Det. directly as the product of two or more Det. ! Still It was no use. I finally concluded that the Second Det. Can't be expressed as a product without Rearrangement ! But Rearranging it makes it more creepier!
Could you please Guide me how to Rearrange it so that it could be easily expressed as a product? Or is their Another Way for this type of Problem? Please Let me know!
Any help would be Appreciated.
 A: Notice that
$$
\begin{bmatrix}
a & b & c \\ 
c & a & b \\ 
b & c & a%
\end{bmatrix}%
^{-1}=\frac{1}{\Delta }%
\begin{bmatrix}
a^{2}-bc & c^{2}-ab & b^{2}-ac \\ 
b^{2}-ac & a^{2}-bc & c^{2}-ab \\ 
c^{2}-ab & b^{2}-ac & a^{2}-bc%
\end{bmatrix}.$$
Then, since $\det(A^{-1})=\det(A)^{-1}$ and $\det(cA)=c^{n}\det(A)$ for a matrix $A$ that is $n\times n$, taking determinants of both sides we get
$$
\frac{1}{\Delta}=\frac{1}{\Delta ^3}%
\det\left(\begin{bmatrix}
a^{2}-bc & c^{2}-ab & b^{2}-ac \\ 
b^{2}-ac & a^{2}-bc & c^{2}-ab \\ 
c^{2}-ab & b^{2}-ac & a^{2}-bc%
\end{bmatrix}\right),$$
so
$$
\det\left(\begin{bmatrix}
a^{2}-bc & c^{2}-ab & b^{2}-ac \\ 
b^{2}-ac & a^{2}-bc & c^{2}-ab \\ 
c^{2}-ab & b^{2}-ac & a^{2}-bc%
\end{bmatrix}\right)=\Delta^2.$$
Finally, notice that $\det(A)=\det(A^T)$.
A: Take $A$ the matrix inside the determinant so that $\Delta=\det(A)$. The transpose of the new matrix is the so called classical adjoint matrix of $A$ for which we have:
$$
\mathrm{adj}(A)A=\det(A) I.
$$
Hence the determinant is equal to $\Delta^{k-1}$ where $k$ is the matrix dimension and is equal to $3$ here.
A: To complete the answer given by @mzp, note that the matrix 
$$
\left(\begin{array}{ccc} a^2 - bc & b^2 - ca & c^2 - ab \\ c^2 - ab & a^2 - bc & b^2 - ca \\ b^2 - ca & c^2 - ab & a^2 - bc \end{array}\right) 
$$
is 
$$
\pmatrix{1&0&0\\0&0&1\\0&1&0}\left( \matrix{a^{2}-bc & c^{2}-ab & b^{2}-ac \\ 
b^{2}-ac & a^{2}-bc & c^{2}-ab \\ 
c^{2}-ab & b^{2}-ac & a^{2}-bc} \right)\pmatrix{1&0&0\\0&0&1\\0&1&0}
$$
and that the determinant of $\pmatrix{1&0&0\\0&0&1\\0&1&0}$ is $-1$.
