Question

Solve:
$\left|\begin{array}{ccc}{a}^2+1& ab& ac\\ ab& b^2+1& bc\\ ca& cb& c^2+1\end{array}\right|=1+a^2+b^2+c^2$

Answer 1

Now we have

$\begin{array}{c}=abc\left|\begin{array}{ccc}a+\frac{1}{a}& a& a\\ b& b+\frac{1}{b}& b\\ c& c& c+\frac{1}{c}\end{array}\right|\\ =a^2{b}^2{c}^2\left|\begin{array}{ccc}1+\frac{1}{a^2}& 1& 1\\ 1& 1+\frac{1}{b^2}& 1\\ 1& 1& 1+\frac{1}{c^2}\end{array}\right|\\ =a^2{b}^2{c}^2\left\{\left(1+\frac{1}{a^2}\right)\left(1+\frac{1}{b^2}\right)\left(1+\frac{1}{c^2}\right)-1-\left(1+\frac{1}{c^2}-1\right)+\left(1-1-\frac{1}{b^2}\right)\right\}\\ =a^2{b}^2{c}^2\left[\left(1+\frac{1}{a^2}\right)\left\{1+\frac{1}{c^2}+\frac{1}{b^2}+\frac{1}{b^2{c}^2}-1\right\}-\frac{1}{c^2}-\frac{1}{b^2}\right]\\ =a^2{b}^2{c}^2\left(\frac{1}{c^2}+\frac{1}{b^2}+\frac{1}{b^2{c}^2}+\frac{1}{a^2{c}^2}+\frac{1}{a^2{b}^2}+\frac{1}{a^2{b}^2{c}^2}-\frac{1}{c^2{b}^2}\right)\\ =a^2{b}^2+a^2{c}^2+a^2+b^2+c^2+1-a^2{b}^2-a^2{c}^2\\ hence,weget\\ =a^2+b^2+c^2+1\end{array}$