To evaluate the given determinant Question: Evaluate the determinant 
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\ 
c^2a^2 & ca & c+a \\ 
a^2b^2 & ab & a+b \\ 
\end{array}
\right|$
My answer:
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\ 
c^2a^2 & ca & c+a \\ 
a^2b^2 & ab & a+b \\ 
\end{array}
\right|= \left|
\begin{array}{cc} b^2c^2 & bc & c \\ 
c^2a^2 & ca & a \\ 
a^2b^2 & ab & b \\ 
\end{array}
\right| + \left|
\begin{array}{cc} b^2c^2 & bc & b \\ 
c^2a^2 & ca & c \\ 
a^2b^2 & ab & a \\ 
\end{array}
\right|= abc \left|
 \begin{array}{cc} bc^2 & c & 1 \\ 
ca^2 & a & 1 \\ 
ab^2 & b & 1 \\ 
\end{array}
\right| +abc \left|
\begin{array}{cc} b^2c & b & 1 \\ 
c^2a & c & 1 \\ 
a^2b & a & 1 \\ 
\end{array}
\right|$
how do I proceed from here?

$F=\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\ 
c^2a^2 & ca & c+a \\ 
a^2b^2 & ab & a+b \\ 
\end{array}
\right|$
$=\dfrac1{abc}\left|
\begin{array}{cc} ab^2c^2 & abc & a(b+c) \\ 
c^2a^2b & bca & b(c+a) \\ 
a^2b^2c & abc & c(a+b) \\ 
\end{array}
\right|$
$=\left|
\begin{array}{cc} ab^2c^2 &1& a(b+c) \\ 
c^2a^2b &1& b(c+a) \\ 
a^2b^2c &1& c(a+b) \\ 
\end{array}
\right|$
$R_3'=R_3-R_1,R_2'=R_2-R_1$
$F=\left|
\begin{array}{cc} ab^2c^2 &1& a(b+c) \\ 
abc^2(a-b) &0& -a(a-b) \\ 
-ab^2c(c-a) &0& b(c-a) \\ 
\end{array}
\right|$
$=(a-b)(c-a)\left|
\begin{array}{cc} ab^2c^2 &1& a(b+c) \\ 
abc^2 &0& -a \\ 
-ab^2c &0& b \\ 
\end{array}
\right|$
$=(a-b)(c-a)(-1)^{1+2}\cdot\left|
\begin{array}{cc} abc^2 & -a \\ 
-ab^2c & b \\ 
\end{array}
\right|$
Can you take it from here?
