If a,b,c are three distinct real numbers If $a,b,c$ are three distinct real numbers and
$$a+\frac1b=b+\frac1c=c+\frac1a=t$$
for some real number $t$ prove that $abc+t=0$
 A: We can use $c = t - 1/a$ to eliminate $c$ from the set of three equations. We obtain:
$$
(t - b)  (t - 1/a) = 1
\\
a = t - 1/b
$$
Using the second formula to eliminate $a$ from the first yields:
$$
t^3 - t^2  (b + 1/b) - t + (b + 1/b) = 0
$$
This third order equation in $t$ can be rewritten as follows.
$$
(t + 1)  (t - 1)  (t - b - 1/b) = 0
$$
We see that $t$ has three solutions: $t = 1$, $t = -1$ and $t = b + 1/b.$ 
(I) $t = 1$. This leads to the solution: $a = x$, $b = 1/(1-x)$, $c = (x-1)/x$ with $x$ a real number in $(-\infty, +\infty)$. The product $abc$ equals $-1$, hence the solution is in agreement with $abc + t = 0$. 
(II) $t = -1$. This leads to the solution: $a = x$, $b = -1/(1+x)$, $c = -(1+x)/x$. Again $x$ is a real number in $(-\infty, +\infty)$. The product $abc$ equals $+1$. Since $t = -1$, in the solution is in agreement with $abc + t = 0$. 
(III) $t = b + 1/b$. This leads to the solution: $a = x$, $b = x$, $c = x$, with $x$ a real number in $(-\infty, +\infty)$. The product $abc$ equals $x^3$. Since $t = x + 1/x$, this solution is not in agreement with $abc + t = 0$. However, the problem states that $a$, $b$ and $c$ must be distinct. On that ground we are forced to omit this solution.  
