Let $f(x)=x^2-kx$ and $g(x)=x^3-kx$, $k\in\mathbb{R^+}$ be two real valued functions such that for a rational number $\alpha$, $f(\alpha)$ and $g(\alpha)$ both are rational numbers.
Find the range of values that $k$ can take.
My Attempt
If $k=1$ then given condition is trivially true. But can there be other values.
If $m$ and $n$ be rational numbers such that $\alpha^2-k\alpha=m$ and $\alpha^3-k\alpha=n$.
This would imply that $k$ is rational since $\alpha,m$ and $n$ are all rational
Also by eliminating $k$, we have $\alpha^2(\alpha-1)=n-m$.
But after this I reach some kind of dead end.
The statement is a bit unclear. I think that you are saying that for a fixed rational $\alpha$, we have $f(\alpha)$ and $g(\alpha)$ are both rational. Then what possible values of $k$ are there (depending possibly on $\alpha$)?
That is, we know only that
\begin{align} \alpha^2 - k \alpha &= a \in \mathbb{Q} \\ \alpha^3 - k \alpha &= b \in \mathbb{Q}. \end{align}
If $\alpha = 0$, then $k$ can be anything and this is uninteresting. Suppose now that $\alpha \neq 0$. Then we can divide by $\alpha$ and find
\begin{align} \alpha - k &= a' \in \mathbb{Q} \\ \alpha^2 - k &= b' \in \mathbb{Q}. \end{align}
The first equation tells us that $k = \alpha - a' \in \mathbb{Q}$, and thus $k$ must be rational. Clearly any rational $k$ suffices.
We gain no new information from $g$, and we conclude that $k$ can be any rational when $\alpha \neq 0$, and otherwise $k$ has no restrictions.