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.

We can assume $\,abc\neq 0\,$ , otherwise the determinant is zero at once (why? For example, suppose $\,b=0\,$ . Then either $\,ac=0\,$ and we get a row of zeros, or the 3rd row is a multiple of the 1st one...and likewise if $\,a=0\vee c=0\,$ ):
$$\begin{vmatrix}b^2c^2 &bc& b+c\cr c^2a^2&ca&c+a\cr a^2b^2&ab&a+b\cr\end{vmatrix}\stackrel{aR_1\,,\,bR_2\,,\,cR_3}{\longrightarrow}\begin{vmatrix}ab^2c^2 &abc& ab+ac\cr a^2bc^2&abc&ab+bc\cr a^2b^2c&abc&ac+bc\cr\end{vmatrix}\stackrel{R_2-R_1\,,\,R_3-R_1}\longrightarrow$$
$$\begin{vmatrix}ab^2c^2 &abc& ab+ac\cr abc^2(a-b)&0&-c(a-b)\cr ab^2c(a-c)&0&-b(a-c)\cr\end{vmatrix}\stackrel{\text{develop by 2nd column}}\longrightarrow$$
$$-abc\begin{vmatrix}abc^2(a-b)&-c(a-b)\\ab^2c(a-c)&-b(a-c)\end{vmatrix}$$
And now factor out $\,c(a-b)\,\,,\,\,b(a-c)\,$ from first and second row resp. , and get
$$-ab^2c^2(a-b)(a-c)\begin{vmatrix}abc&-1\\abc&-1\end{vmatrix}=0$$
Finally, pay attention to the fact that the original determinant is zero (or not) without any regard to the above operations
