Show that `|1a b+c1b c+a1c a+b|=0` .

Answer

Proved

Explanation

We know that for a `3 \times 3` matrix $$A=\left[\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right]$$ the determinant is:
$$ |A|=a(e i-f h)-b(d i-f g)+c(d h-e g) $$
Here, the given matrix is $$\mathrm{A}=\left[\begin{array}{ccc}1 & \mathrm{a} & \mathrm{b}+\mathrm{c} \\ 1 & \mathrm{~b} & \mathrm{c}+\mathrm{a} \\ 1 & \mathrm{c} & \mathrm{a}+\mathrm{b}\end{array}\right]$$ Then, the the determinant of `A` is:
$$
\begin{aligned}
&|\mathrm{A}|=1[\mathrm{~b}(\mathrm{a}+\mathrm{b})-\mathrm{c}(\mathrm{c}+\mathrm{a})]-\mathrm{a}[1(\mathrm{a}+\mathrm{b})-1(\mathrm{c}+\mathrm{a})]+(\mathrm{b}+\mathrm{c})[1(\mathrm{c})-1(\mathrm{~b})] \\
...