You seem to be asking for a proof of an obviously contradictory statement. So are you really just looking for a proof that $x+yt+zt^2\neq 0$ for all $x,y,z\in \mathbb{Q}$ not all zero?

JoeBlow wrote: You seem to be asking for a proof of an obviously contradictory statement. So are you really just looking for a proof that $x+yt+zt^2\neq 0$ for all $x,y,z\in \mathbb{Q}$ not all zero? Sorry, edited.

WakeUp wrote: Prove that if $x+yt+zt^2=0$... The point is this situation is already impossible unless $x=y=z=0$. So any statement that starts out with these words is true (vacuously so). For example, it is true that if $x,y,z\in \mathbb{Q}$, $x+yt+zt^2=0$ and $x,y,z$ are not all $0$, then all pink elephants can fly. Indeed, suppose $x+yt+zt^2=0$ for some $x,y,z\in \mathbb{Q}$ with $z\neq 0$, then solving the quadratic for $t$ we find $t=p+\sqrt{q}$ for some $p,q\in \mathbb{Q}$. This implies $2=p^3+3pq+(3p^2+q)\sqrt{q}$, and hence either $\sqrt{q}\in \mathbb{Q}$ or $3p^2+q=0$. $\sqrt{q}\in \mathbb{Q}$ implies $t\in \mathbb{Q}$, and $3p^2+q=0$ implies $t\not \in \mathbb{R}$, either way we have a contradiction.

The last time I saw this very same question asked, the condition was that $x + yt + zt^2 \neq 0$. Typo on the OP's part.

Apologies, again, Mellow Melon is right and the condition should be $x+yt+zt^{2}\not= 0$. If I'm posting problems for a whole Olympiad (e.g. this is Ibero 88), I do not have time to think about every question.

I think the solution can be generalized to any "t" that is the cubic root of a positive integer, so I'll simply use $t=\sqrt[3]k $ Expanding the original condition yields: $ (xu + kyw + kzv) + t(xv + yu + kzw) + t^2 (xw + yv + zu) = 1$ A solution can be found if the following system of equations has a solution for every rational $x,y,z$ $\left. \begin{array}{rcl} xu + kzv + kyw & = & 1 \\ yu + xv + kzw & = & 0 \\ zu + yv + xw & = & 0 \end{array} \right\}$ But this system has rational solutions if and only if its determinant is different to zero (clearly, we are considering $u,v,w$ as variables, and $k,x,y,z$ as given constants). So suppose the determinant equals zero, then, we get: $ x (x^2 - kyz) - kz (xy - kz^2) + ky (y^2 - xz) = 0$ $\leftrightarrow x^3 + ky^3 + k^2z^3 = 3kxyz$ Now write $ x= \dfrac{a}{b} , y= \dfrac{c}{d}, z= \dfrac{e}{f}$, where $a,b,c,d,e,f$ are integers since $x, y, z$ are rational. Then: $ \dfrac{a^3}{b^3} + k\dfrac{c^3}{d^3} + k^2\dfrac{e^3}{f^3} = 3\dfrac{ace}{bdf}$ Multiplying by $(bdf)^3$ and simplyfing: $ (adf)^3 + k(cbf)^3 + k^2(ebd)^3 = 3kabc(bdf)^2$ Now taking $p=adf, q=cbf, r= ebd$: $p^3 + kq^3 + k^2r^3 = 3kpqr$ From here it is pretty straightforward. Suppose an integer solution $(p,q,r)$ whose absolute values are minimal. But clearly k divides $p^3$, so $k^3$ divides $p^3$. Let $p= kp'$, then: $p'^3k^3 + kq^3 + k^2r^3 = 3k^2p'qr$ Simplyfing : $p'^3k^2 + q^3 + kr^3 = 3kp'qr$ , and $k$ divides $q^3$. Repeating with $r$ the same procedure we get $(\dfrac{p}{k},\dfrac{q}{k},\dfrac{r}{k})$ is another integer solution, contradicting the minimality first conejctured. So the determinant is always different to zero, and rational solutions $(u,v,w)$ always exist.