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  1. #1
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    three distinct

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

    a + (1 / b)  =b + (1 / c)  =c + (1 / a)  =t<br />

    for some real number t

    prove that  abc + t=0 .
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  2. #2
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    Quote Originally Posted by perash View Post
    If a,b,c are three distinct nonzero real numbers and

    a + (1 / b) =b + (1 / c) =c + (1 / a) =t<br />

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

    a + (1 / b)  =b + (1 / c)  =c + (1 / a)  =t<br />

    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
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