system of equations

equilibrium · New Member Offline Joined: 28 Oct 2009 Posts: 3

x^3 - 3x^2 + 5x = 1

y^3 - 3y^2 + 5y = 5

Find x + y = ?

**Posted:** Wed Oct 28, 2009 3:23 am

Spribo · Poincare Conjecture Offline Joined: 06 Jul 2008 Posts: 104

I use the fact that $k^2 - kl + l^2\geq 0$ .
$x^3 - 3x^2 + 5x = 1, x^3 - 3x^2 + 3x - 1 = - 2x, (x - 1)^3 = - 2x$ . Similarly $(y - 1)^3 = 4 - 2y$ . Set $x - 1 = k$ , $y - 1 = l$ , $x + y = a$ . Adding the two previous equations you get $(a - 2)(k^2 - kl + l^2) = 4 - 2a$ . First solution: $a = 2$ . If $a\neq 2$ , then $k^2 - kl + l^2 = - 2$ , which is impossible. So, $x + y = 2$ .

**Posted:** Wed Oct 28, 2009 11:47 am

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equilibrium · New Member Offline Joined: 28 Oct 2009 Posts: 3

thats right, thank you!

**Posted:** Wed Oct 28, 2009 11:43 pm

mrmohamed

**Posted:** Sat Nov 07, 2009 2:36 pm

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