Without expanding, how to prove that

Question

Without expanding, how to prove that

$$\begin{vmatrix} bc & ab & a \\ ac & b^2 & b \\ b & c & 1 \end{vmatrix} = (a+b) \begin{vmatrix} 1 & b & 1 \\ a & b^2 & b \\ b & c^2 & c \end{vmatrix} ?$$

I've tried to sum the rows or columns but no progress.

Answer 1

On the left hand side, we have$\begin{vmatrix} bc & ab & 1 \\ ac & b^2 & b \\ b & c & a \end{vmatrix} = bc \begin{vmatrix} b^2 & b \\ c & a \end{vmatrix} -ab \begin{vmatrix} ac & b \\ b & a \end{vmatrix} + \begin{vmatrix} ac & b^2 \\ b & c \end{vmatrix} = bc(ab^2-bc)-ab(a^2c-b^2) + (ac^2-b^3) = ab^3c-b^2c^2-a^3bc+ab^3+ac^2-b^3.$

On the right hand side, we have $(a+b)\begin{vmatrix} 1 & b & 1 \\ a & b^2 & b \\ b & c^2 & c \end{vmatrix} = (a+b)\left( \begin{vmatrix} b^2 & b \\ c^2 & c \end{vmatrix} -b \begin{vmatrix} a & b \\ b & c \end{vmatrix} + \begin{vmatrix} a & b^2 \\ b & c^2 \end{vmatrix} \right)= (a+b)((b^2c-bc^2)-b(ac-b^2)+(ac^2-b^3)) = (a+b)(b^2c-bc^2-bac+b^3+ac^2-b^3) = -a^2bc+a^2c^2+b^3c-b^2c^2.$

This leads me to believe the problem is stated incorrectly.