If $\begin{vmatrix} -1 & a & a \\ b & -1 & b \\ c & c & -1 \end{vmatrix} =0$ then what's the value of $$\frac{1}{1+a}+\frac{1}{1+b} +\frac{1}{1+c}$$
I just expanded the Determinant, to get $$ab+bc+ac+2abc=1$$
Which further leads to $$\frac{1+a}{a}+\frac{1+b}{b}+ \frac{1+c}{c}= 1+\frac{1}{abc}$$
The solution in the book uses elementary operations, but is it possible to make the "original" equation of determinant in the required form?
This maybe a duplicate question, but I couldn't find it.
With $$c=\frac{1-ab}{a+b+2ab}$$ we get $$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+\frac{1-ab}{a+b+2ab}}=2$$ after a few simplifications.