<s>[INST] 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.
 [/INST] \begin{align*}
\begin{vmatrix}b^2c^2&bc&b+c\\c^2a^2&ca&c+a\\a^2b^2&ab&a+b\end{vmatrix}
\stackrel{\large R_1-R_3\,,\,R_2-R_3}{\longrightarrow}
&(c-a)(c-b)
\,\begin{vmatrix}b^2(c+a)&b&1\\a^2(c+b)&a&1\\a^2b^2&ab&a+b\end{vmatrix}\\
\stackrel{\large C_1\,-\,ab\,C_2}{\longrightarrow}
&(c-a)(c-b)c
\,\begin{vmatrix}b^2&b&1\\a^2&a&1\\0&ab&a+b\end{vmatrix}\\
\stackrel{\large R_1-R_2}{\longrightarrow}
&(c-a)(c-b)c(b-a)
\,\begin{vmatrix}b+a&1&0\\a^2&a&1\\0&ab&a+b\end{vmatrix}\\
\stackrel{\large R_2-aR_1}{\longrightarrow}
&(c-a)(c-b)c(b-a)
\,\begin{vmatrix}b+a&1&0\\-ab&0&1\\0&ab&a+b\end{vmatrix}=0.
\end{align*}
</s>