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(D) Given the determinant equation: $\left|\begin{array}{ccc}1+x & 1 & 1 \ 1 & 1+y & 1 \ 1 & 1 & 1+z\end{array}ight|=0$ Divide the first row by $x

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$-1$

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(D) Given the determinant equation: $\left|\begin{array}{ccc}1+x & 1 & 1 \ 1 & 1+y & 1 \ 1 & 1 & 1+z\end{array}ight|=0$ Divide the first row by $x$,the second row by $y$,and the third row by $z$: $xyz \left|\begin{array}{ccc}\frac{1}{x}+1 & \frac{1}{x} & \frac{1}{x} \ \frac{1}{y} & \frac{1}{y}+1 & \frac{1}{y} \ \frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{array}ight|=0$ Since $x, y, z 
eq 0$,we can divide by $xyz$: $\left|\begin{array}{ccc}\frac{1}{x}+1 & \frac{1}{x} & \frac{1}{x} \ \frac{1}{y} & \frac{1}{y}+1 & \frac{1}{y} \ \frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{array}ight|=0$ Apply the operation $C_1 ightarrow C_1 + C_2 + C_3$: $\left|\begin{array}{ccc}1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & \frac{1}{x} & \frac{1}{x} \ 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & \frac{1}{y}+1 & \frac{1}{y} \ 1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} & \frac{1}{z} & \frac{1}{z}+1\end{array}ight|=0$ Factor out $(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$: $(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \left|\begin{array}{ccc}1 & \frac{1}{x} & \frac{1}{x} \ 1 & \frac{1}{y}+1 & \frac{1}{y} \ 1 & \frac{1}{z} & \frac{1}{z}+1\end{array}ight|=0$ Since $x, y, z$ are distinct,the determinant part is non-zero. Therefore,$1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = 0$ $x^{-1}+y^{-1}+z^{-1} = -1$

If $x, y, z$ are all different and not equal to zero and $\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0$,then the value of $x^{-1}+y^{-1}+z^{-1}$ is equal to

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