Let $ f(t)=t^3+xt+y$. Then $ f(a)=f(b)=f(c)=0$. So $ f(t)=(t-a)q(x)$. Now $ f(b)=(b-a)q(b)=0$ but $ b-a\ne 0$ so $ q(b)=q(c)=0$. So $ q(t)=(t-b)r(t)$ and $ r(c)=0$ similarly so $ f(t)=(t-a)(t-b)(t-c)$. Expanding, we get $ a+b+c=0$.

Altheman wrote: Let $ f(t) = t^3 + xt + y$. Then $ f(a) = f(b) = f(c) = 0$. So $ f(t) = (t - a)q(x)$. Now $ f(b) = (b - a)q(b) = 0$ but $ b - a\ne 0$ so $ q(b) = q(c) = 0$. So $ q(t) = (t - b)r(t)$ and $ r(c) = 0$ similarly so $ f(t) = (t - a)(t - b)(t - c)$. Expanding, we get $ a + b + c = 0$. No need for so much explanation, it's known that $ f(t)=g(t)(t-a)(t-b)(t-c)$, and as $ f(t)$ is of third degree, $ g(t)=1$

Kill it in 3 seconds by Vieta.

Bugi wrote: Altheman wrote: Let $ f(t) = t^3 + xt + y$. Then $ f(a) = f(b) = f(c) = 0$. So $ f(t) = (t - a)q(x)$. Now $ f(b) = (b - a)q(b) = 0$ but $ b - a\ne 0$ so $ q(b) = q(c) = 0$. So $ q(t) = (t - b)r(t)$ and $ r(c) = 0$ similarly so $ f(t) = (t - a)(t - b)(t - c)$. Expanding, we get $ a + b + c = 0$. No need for so much explanation, it's known that $ f(t) = g(t)(t - a)(t - b)(t - c)$, and as $ f(t)$ is of third degree, $ g(t) = 1$ I think that we have to be careful. If $ a=b$ or something we do not have that result.

Altheman wrote: Bugi wrote: Altheman wrote: Let $ f(t) = t^3 + xt + y$. Then $ f(a) = f(b) = f(c) = 0$. So $ f(t) = (t - a)q(x)$. Now $ f(b) = (b - a)q(b) = 0$ but $ b - a\ne 0$ so $ q(b) = q(c) = 0$. So $ q(t) = (t - b)r(t)$ and $ r(c) = 0$ similarly so $ f(t) = (t - a)(t - b)(t - c)$. Expanding, we get $ a + b + c = 0$. No need for so much explanation, it's known that $ f(t) = g(t)(t - a)(t - b)(t - c)$, and as $ f(t)$ is of third degree, $ g(t) = 1$ I think that we have to be careful. If $ a = b$ or something we do not have that result. It says they are distinct (look at Valentin's post)

Well of course. The point is to link that hypothesis to the fact that there is a factorization. It seems to me that if you are actually going to prove something in this problem (as opposed to trivializing it by citing some theorem) you use the division algorithm to write that representation of the polynomial.

Valentin Vornicu wrote: Let $ a,b,c,x,y$ be five real numbers such that $ a^3 + ax + y = 0$, $ b^3 + bx + y = 0$ and $ c^3 + cx + y = 0$. If $ a,b,c$ are all distinct numbers prove that their sum is zero. Ciprus $ (a^3+ax+y)-(b^3+bx+y) = 0 \iff (a-b)(a^2+ab+b^2+x)=0$ $ \iff a^2+ab+b^2+x=0$ (since $ a \neq b$) and likewise $ b^2+bc+c^2+x=0$ since $ b \neq c$. Now $ (a^2+ab+b^2+x)-(b^2+bc+c^2+x)=0 \iff$ $ a^2-c^2+b(a-c) = (a-c)(a+b+c) = 0 \iff a+b+c = 0$ (since $ a \neq c$)

Mathias_DK wrote: Valentin Vornicu wrote: Let $ a,b,c,x,y$ be five real numbers such that $ a^3 + ax + y = 0$, $ b^3 + bx + y = 0$ and $ c^3 + cx + y = 0$. If $ a,b,c$ are all distinct numbers prove that their sum is zero. Ciprus $ (a^3 + ax + y) - (b^3 + bx + y) = 0 \iff (a - b)(a^2 + ab + b^2 + x) = 0$ $ \iff a^2 + ab + b^2 + x = 0$ (since $ a \neq b$) and likewise $ b^2 + bc + c^2 + x = 0$ since $ b \neq c$. Now $ (a^2 + ab + b^2 + x) - (b^2 + bc + c^2 + x) = 0 \iff$ $ a^2 - c^2 + b(a - c) = (a - c)(a + b + c) = 0 \iff a + b + c = 0$ (since $ a \neq c$) Isn't it the same as qntty's solution?

Bugi wrote: Mathias_DK wrote: Valentin Vornicu wrote: Let $ a,b,c,x,y$ be five real numbers such that $ a^3 + ax + y = 0$, $ b^3 + bx + y = 0$ and $ c^3 + cx + y = 0$. If $ a,b,c$ are all distinct numbers prove that their sum is zero. Ciprus $ (a^3 + ax + y) - (b^3 + bx + y) = 0 \iff (a - b)(a^2 + ab + b^2 + x) = 0$ $ \iff a^2 + ab + b^2 + x = 0$ (since $ a \neq b$) and likewise $ b^2 + bc + c^2 + x = 0$ since $ b \neq c$. Now $ (a^2 + ab + b^2 + x) - (b^2 + bc + c^2 + x) = 0 \iff$ $ a^2 - c^2 + b(a - c) = (a - c)(a + b + c) = 0 \iff a + b + c = 0$ (since $ a \neq c$) Isn't it the same as qntty's solution? Yes it is.

leepakhin wrote:

Would this get a good score on an actual olympiad?

tanuagg13 wrote: leepakhin wrote:

Would this get a good score on an actual olympiad? You would have to elaborate more.

a,b,c is the roots of the one equation k*k*k+k*x+y=0 By Vieta a+b+c=0