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?
 A: This is probably the shortest way:-
$$
\begin{align}
\begin{vmatrix} b^2c^2 & bc & b+c \\\ 
c^2a^2 & ca & c+a \\\ 
a^2b^2 & ab & a+b 
\end{vmatrix}
 &=a^3b^3c^3\begin{vmatrix}a^{-1} &1 & b^{-1}+c^{-1}\\\ b^{-1} &1&a^{-1}+c^{-1}\\\ c^{-1} &1& a^{-1}+b^{-1}\end{vmatrix}\\
&=(abc)^3\left(\frac 1a+\frac 1b+\frac 1c\right)\begin{vmatrix}
a^{-1}&1&1\\\ b^{-1}&1&1\\\ c^{-1} &1 &1
\end{vmatrix}
\\
&=0
\end{align}
$$
What I did was to first take out $bc$ , $ca$ and $ab$ common from the rows individually. Then I took out $abc$ from the first column.
A: $F=\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\ 
c^2a^2 & ca & c+a \\ 
a^2b^2 & ab & a+b \\ 
\end{array}
\right|$
$=\dfrac1{abc}\left|
\begin{array}{cc} ab^2c^2 & abc & a(b+c) \\ 
c^2a^2b & bca & b(c+a) \\ 
a^2b^2c & abc & c(a+b) \\ 
\end{array}
\right|$
$=\left|
\begin{array}{cc} ab^2c^2 &1& a(b+c) \\ 
c^2a^2b &1& b(c+a) \\ 
a^2b^2c &1& c(a+b) \\ 
\end{array}
\right|$
$R_3'=R_3-R_1,R_2'=R_2-R_1$
$F=\left|
\begin{array}{cc} ab^2c^2 &1& a(b+c) \\ 
abc^2(a-b) &0& -a(a-b) \\ 
-ab^2c(c-a) &0& b(c-a) \\ 
\end{array}
\right|$
$=(a-b)(c-a)\left|
\begin{array}{cc} ab^2c^2 &1& a(b+c) \\ 
abc^2 &0& -a \\ 
-ab^2c &0& b \\ 
\end{array}
\right|$
$=(a-b)(c-a)(-1)^{1+2}\cdot\left|
\begin{array}{cc} abc^2 & -a \\ 
-ab^2c & b \\ 
\end{array}
\right|$
Can you take it from here?
A: Your given matrix is : $\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\ 
c^2a^2 & ca & c+a \\ 
a^2b^2 & ab & a+b \\ 
\end{array}
\right|$
Determinant of the given matrix is :
$\implies b^2c^2[ca^2 + abc - abc + a^2b] - bc[a^3c^2 + a^2bc^2 - a^2b^2c - a^3b^2] + (b+c)[a^3bc^2 - a^3b^2c]$
$\implies b^2c^2[ca^2  + a^2b] - bc[a^3c^2 + a^2bc^2 - a^2b^2c - a^3b^2] + (b+c)[a^3bc^2 - a^3b^2c]$
$\implies a^2b^2c^3 - a^2b^3c^2 - a^3bc^3 - a^2b^2c^3 + a^2b^3c^2 + a^3b^3c + a^3b^2c^2 - a^3b^3c + a^3bc^3 - a^3b^2c^2$
$\implies a^2b^2c^3 - a^2b^2c^3 - a^2b^3c^2 + a^2b^3c^2 - a^3bc^3 + a^3bc^3  + a^3b^3c - a^3b^3c + a^3b^2c^2  - a^3b^2c^2$
$\implies 0$
Therefore, given matrix is singular.
