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If x and y be real numbers such that, $x^3 - 3x^2 + 5x = 1$ and $y^3 - 3x^2 + 5y = 5$; Find $(x + y)$ From an old Russian olympiad.

I tried to make the equations homogenous by substituting for $1 = x^3 - 3x^2 + 5x$ in the second equation for $5 * 1$, which didn't work even by repeated substitution.

Next, equating $3x^2$ in both equations and trying to homogenise them also doesn't give me anything.

Hint for this one please!

DeepSea
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buzaku
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  • Have you tried using the formula for the roots of a cubic on the left equation, and then substituting back into the right one? Brutal but should work – Gabriel Mar 05 '15 at 05:09
  • The discriminant for the cubic is negative, so it has one real root and the other 2 complex conjugates. The formulae for roots of a cubic just get very complicated from here on. I feel, there is an elegant symmetry in the problem which is suggesting avoidance of brute force. :) – buzaku Mar 05 '15 at 05:52
  • Since it is from an olympiad, I assume the answer should be an integer? – Vincenzo Oliva Mar 05 '15 at 08:32

1 Answers1

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I think you have somewrong,I have see a book with Russian olympiad problem

Let $x,y$ are real numbers, and $$x^3-3x^2+5x=1,y^3-\color{#0a0}{\text{$3y^2$}}+5y=5$$ Find $x+y$

since $$(x-1)^3+2(x-1)=-2$$ $$(y-1)^3+2(y-1)=2$$ since $f(x)=x^3+2x$ is odd function,and increaing on $R$ since $$f(x-1)+f(y-1)=0\Longleftrightarrow f(x-1)=-f(y-1)=f(1-y)$$ so $$x-1=1-y\Longrightarrow x+y=2$$

badskjapanses
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    I guess this might be the solution, since otherwise the equations seems require just brute force. No elegant way out. It would be sad if this question was difficult because of a damn print error in my problem book. – buzaku Mar 05 '15 at 07:01