13.424 Three machines perform a certain job. If the first one does the job alone, it will finish $a$ days later than if all machines work together. If the second one does the job alone, it will finish $b$ days later than all together, and if the third one does the job alone, it will take $c$ times longer than all machines working together. How many days does each of them take to complete the job individually? What values can $c$ take?
Solution. Let $x, y, z$ be the number of days it takes for the first, second, and third machines, respectively, to complete the entire work $V$ individually. Then $\frac{V}{x}, \frac{V}{y}, \frac{V}{z}$ are the efficiencies of the machines, $\frac{z}{c}$ is the time it takes for them to complete the entire work together. Then

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=\frac{z}{c}+a, \\
y=\frac{z}{c}+b, \\
V=\frac{z}{c}\left(\frac{V}{x}+\frac{V}{y}+\frac{V}{z}\right)
\end{array} \Rightarrow 1=\frac{z}{c}\left(\frac{c}{z+a c}+\frac{c}{z+b c}+\frac{1}{z}\right) \Rightarrow\right. \\
& \Rightarrow(c+1) z^{2}+z(a c+b c)+a b c^{2}-a b c^{3}=0 \Rightarrow z=\frac{-c(a+b)+c \sqrt{(a-b)^{2}+4 a b c^{2}}}{2(c+1)}
\end{aligned}
$$

Since $z>0$, then $\sqrt{(a-b)^{2}+4 a b c^{2}}>a+b \Leftrightarrow c^{2}>1 \Leftrightarrow c>1$.

Answer: $\quad a+\frac{-(a+b)+\sqrt{(a-b)^{2}+4 a b c^{2}}}{2(c+1)} ; b+\frac{-(a+b)+\sqrt{(a-b)^{2}+4 a b c^{2}}}{2(c+1)}$;

$$
\frac{-c(a+b)+c \sqrt{(a-b)^{2}+4 a b c^{2}}}{2(c+1)}, \text { where } c>1 \text {. }
$$