1. ## three distinct

If a,b,c are three distinct nonzero real numbers and

$a + (1 / b) =b + (1 / c) =c + (1 / a) =t
$

for some real number t

prove that $abc + t=0$ .

2. Originally Posted by perash
If a,b,c are three distinct nonzero real numbers and

$a + (1 / b) =b + (1 / c) =c + (1 / a) =t
$

for some real number t

prove that $abc + t=0$ .

so we have: $ab+1=bt, \ \ bc+ 1 = ct, \ \ ac + 1 = at,$ 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:

$(a-c)b=(b-c)t, \ \ (b-a)c=(c-a)t, \ \ (c-b)a=(a-b)t,$ and multiply the equations together:

$(b-a)(c-b)(a-c)abc=(a-b)(b-c)(c-a)t^3,$ which, since $a \neq b \neq c,$ gives us: $abc+t^3=0. \ \ \ \ (*)$

this time multilpy 1) by $c$, 2) by $a$, and 3) by $b$ to get: $abc+c=bct, \ \ abc+a=act, \ \ abc+b=abt,$

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: $c-a=(b-a)ct, \ \ a-b=(c-b)at, \ \ b-c=(a-c)bt,$ and multiply the equations together:

$(a-b)(b-c)(c-a)=(b-a)(c-b)(a-c)abct^3,$ which, since $a \neq b \neq c,$ gives us: $abct^3 = -1. \ \ \ \ \ (**)$

now by $(*)$ and $(**): \ t^6 = 1.$ thus: $t = \pm 1,$ since $t \in \mathbb{R}.$ so: $t^3=t,$ and hence by $(*): \ \ abc+t=0. \ \ \ \square$

3. Originally Posted by perash
If a,b,c are three distinct nonzero real numbers and

$a + (1 / b) =b + (1 / c) =c + (1 / a) =t
$

for some real number t

prove that $abc + t=0$ .
I found an alternative, simpler way:

$ab+1 = bt, bc+1 = ct, ac+1 = at$

$abc+c = bct, abc+a = act, abc+b = abt$

$abc + t= bct+t - c = act+t - a = abt+t - b$

$abc + t= ct^2 - c = at^2 - a = bt^2 - b$

$ct^2 - c = at^2 - a \Rightarrow (t^2 - 1)(c-a) = 0 \Rightarrow t^2 = 1$

This implies $abc + t= ct^2 - c = c(t^2 - 1) = 0$