TITLE: Determinant proof using its properties
QUESTION [5 upvotes]: Prove without expanding:
\begin{equation}
\begin{vmatrix}bc&a^2&a^2\\b^2&ac&b^2\\c^2&c^2 & ab\end{vmatrix} = \begin{vmatrix}ac&bc&ab\\bc&ab&ac\\ab&ac&bc\end{vmatrix}
\end{equation}

Tried to multiply by 'abc' in all rows then take common factors.
Tried to expand determinant into two determinants. 
I used the determinant properties shown here: http://www.vitutor.com/alg/determinants/properties_determinants.html

REPLY [5 votes]: We check that the property holds when one of $a,b,c$ is zero.  Now assume $a,b,c$ are all non-zero.
We divide the columns by $bc$, $ac$, and $ab$, respectively, and multiply the rows by $bc$, $ac$, and $ab$ respectively.  This gives
$$\begin{vmatrix}bc&a^2&a^2\\b^2&ac&b^2\\c^2&c^2 & ab\end{vmatrix} = \begin{vmatrix}bc&ab&ac\\ab&ac&bc\\ac&bc&ab\end{vmatrix}$$
then we cycle the rows to obtain $$=\begin{vmatrix}ac&bc&ab\\bc&ab&ac\\ab&ac&bc\end{vmatrix}.$$