If a,b,c are three distinct nonzero real numbers and
$\displaystyle a + (1 / b) =b + (1 / c) =c + (1 / a) =t
$
for some real number t
prove that $\displaystyle abc + t=0$ .
so we have: $\displaystyle ab+1=bt, \ \ bc+ 1 = ct, \ \ ac + 1 = at,$ which we will call 1), 2), and 3) respectively.
now first subtract 2) from 1), then 3) from 2), and finally 1) from 3) to get:
$\displaystyle (a-c)b=(b-c)t, \ \ (b-a)c=(c-a)t, \ \ (c-b)a=(a-b)t,$ and multiply the equations together:
$\displaystyle (b-a)(c-b)(a-c)abc=(a-b)(b-c)(c-a)t^3,$ which, since $\displaystyle a \neq b \neq c,$ gives us: $\displaystyle abc+t^3=0. \ \ \ \ (*)$
this time multilpy 1) by $\displaystyle c$, 2) by $\displaystyle a$, and 3) by $\displaystyle b$ to get: $\displaystyle abc+c=bct, \ \ abc+a=act, \ \ abc+b=abt,$
which we'll call 4), 5), and 6) respectively. now subtract 5) from 4), then 6) from 5) and finally 4) from 6)
to get: $\displaystyle c-a=(b-a)ct, \ \ a-b=(c-b)at, \ \ b-c=(a-c)bt,$ and multiply the equations together:
$\displaystyle (a-b)(b-c)(c-a)=(b-a)(c-b)(a-c)abct^3,$ which, since $\displaystyle a \neq b \neq c,$ gives us: $\displaystyle abct^3 = -1. \ \ \ \ \ (**)$
now by $\displaystyle (*)$ and $\displaystyle (**): \ t^6 = 1.$ thus: $\displaystyle t = \pm 1,$ since $\displaystyle t \in \mathbb{R}.$ so: $\displaystyle t^3=t,$ and hence by $\displaystyle (*): \ \ abc+t=0. \ \ \ \square$
I found an alternative, simpler way:
$\displaystyle ab+1 = bt, bc+1 = ct, ac+1 = at$
$\displaystyle abc+c = bct, abc+a = act, abc+b = abt$
$\displaystyle abc + t= bct+t - c = act+t - a = abt+t - b$
$\displaystyle abc + t= ct^2 - c = at^2 - a = bt^2 - b$
$\displaystyle ct^2 - c = at^2 - a \Rightarrow (t^2 - 1)(c-a) = 0 \Rightarrow t^2 = 1$
This implies $\displaystyle abc + t= ct^2 - c = c(t^2 - 1) = 0$