Let $a,b$ and $c$ be real numbers.evaluate the following determinant: |$b^2c^2 ,bc, b+c;c^2a^2,ca,c+a;a^2b^2,ab,a+b$| Let $a,b$ and $c$ be real numbers. Evaluate the following determinant:
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}$$
after long calculation I get that the answer will be $0$. Is there any short processs? Please help someone thank you.

If $b=0,$ 
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix} =\begin{vmatrix}0 &0&c\cr c^2a^2&ca&c+a\cr 0&0&a\cr\end{vmatrix}$$
Now, if $a=0,$ $$\text{the determinant becomes }\begin{vmatrix}0 &0&c\cr 0&0&c\cr 0&0&0\cr\end{vmatrix}=0$$
else for $ca\ne 0$ $$\text{the determinant becomes }c^3a^3\begin{vmatrix}0 &0&c\cr 1&1&c\cr 0&0&0\cr\end{vmatrix}=0$$
So, if $abc\ne 0,$
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}$$
$$=\frac1{abc}\begin{vmatrix}a(b^2c^2) &abc& a(b+c)\cr b(c^2a^2)&cab&b(c+a)\cr c(a^2b^2)&abc&c(a+b)\cr\end{vmatrix} \text{ applying  } R_1'=aR_1, R_2'=bR_2, R_3'=cR_3$$
$$=abc\begin{vmatrix}bc &1& a(b+c)\cr ca&1&b(c+a)\cr ab&1&c(a+b)\cr\end{vmatrix}\text{  applying } C_1'=\frac{C_1}{abc} \text{ and } C_2'=\frac{C_2}{abc}$$.
$$=abc\begin{vmatrix}bc &1& a(b+c)+bc\cr ca&1&b(c+a)+ca\cr ab&1&c(a+b)+ab\cr\end{vmatrix} \text{  applying } C_3'=C_3+C_1$$
If $ab+bc+ca=0,$  $$\text{the determinant becomes } abc\begin{vmatrix}bc &1& 0\cr ca&1&0\cr ab&1&0\cr\end{vmatrix}=0  $$
Else 
$$=abc(ab+bc+ca)\begin{vmatrix}bc &1& 1\cr ca&1&1\cr ab&1&1\cr\end{vmatrix} \text{ 
 applying } C_3'=\frac{C_3}{ab+bc+ca}$$
$$=abc(ab+bc+ca)\cdot 0 \text {   as the 2nd & the last columns are identical.}$$
