Algebraic calculation with polynomial and complex root. 
Let $f=X^{3}-7 X+7$ be in $\mathbb{Q}[X]$.
Let $\alpha \in \mathbb{C}$ be a root of $f$ and hence $1, \alpha, \alpha^{2}$ be a basis of the $\mathbb{Q}$ vector space $\mathbb{Q}(\alpha)$. Let $\beta=3 \alpha^{2}+4 \alpha-14$. Write $\beta^{2}$ and $\beta^{3}$ as linear combinations of $1, \alpha, \alpha^{2}$ over $\mathbb{Q}$ and conclude that $\beta$ is a zero of $f$. Is $\beta=\alpha$?


I found my mistake with your help.
The solution: We have
$$
\beta^2=(3\alpha^{2}+4 \alpha-14)^2=9 \alpha^4 + 24 \alpha^3 - 68 \alpha^2 - 112 \alpha + 196
$$
To remove the cubic and quartic powers, we use the fact that $\alpha^3-7\alpha+7=0$ holds and gradually eliminate partial polynomials.
We get
$$
   \beta^2=9 \alpha^4- 68 \alpha^2 + 24 \alpha^3 + 16\cdot (-7 \alpha) + 28\cdot 7
=9 \alpha^4- 68 \alpha^2 -8\cdot (-7 \alpha) \
+ 4\cdot 7\alpha(9 \alpha^3+10\cdot(-7) \alpha +8\cdot 7+2\alpha)+4\cdot 7=\alpha(-5\alpha -7) + 4\cdot 7=-5\alpha^2-7\alpha+28.
$$
Now we try $\beta^3$ with the same method
$$
\beta^3=\beta^2\beta=(3\alpha^{2}+4 \alpha-14)^2(3\alpha^{2}+4 \alpha-14)=(-5\alpha^2-7\alpha+28)(3\alpha^{2}+4 \alpha-14)\
=-15\alpha^4+ 126 \alpha^2 -(41 \alpha^3  +30\cdot(-7\alpha) + 56\cdot 7)=-15\alpha^4+ 126 \alpha^2 -(  -11\cdot(-7\alpha) + 15\cdot 7)\
=-\alpha(15\alpha^3+18\cdot(-7\alpha)  +7\cdot 11) - 15\cdot 7=-\alpha(3\cdot(-7\alpha)  -4\cdot7) - 15\cdot 7=21\alpha^2  +28\alpha-105.
$$
And with this i get indeed
$$
f(\beta)=0.
$$
Thanks for your help! Maybe there are different reasonings for why $\beta\neq \alpha$?
 A: Your mistake appears to be in the last line evaluating $\beta^3$.  The expression
\begin{eqnarray*}
15\alpha^3+18⋅(−7\alpha)+77 &=& 15(7\alpha-7)-7*18\alpha+7*11\\
& =& 7(15-18)\alpha+7(-15+11) \\
& = & -21\alpha -28
\end{eqnarray*}
whereas you got $-21\alpha+62$.  The expression for $\beta^3$ becomes $-21\alpha^2-28\alpha-105$.  Plugging this new expression into the rest of your derivation shows that $\beta$ is a root of the polynomial.  (btw. I cheated and used sympy to check your calculations...)
A: $\beta$ cannot equal $\alpha$. If so, then $\alpha$ would be a root of the polynomial $3x^2+3x-14$. But your starting polynomial $x^3-7x+7$ is irreducible over $\mathbb{Q}[x]$ (use Eisenstein criterion for instance), so $x^3-7x+7$ would divide $3x^2+3x-14$, which is impossible.
A: You should review your computations, there must be some mistake in them, I tried myself to compute $\beta^{2}$ and $\beta^{3}$ and found
$$
\beta^{2} = -5\alpha^{2}-7\alpha+28,
$$
$$
\beta^{3} = 21\alpha^{2} +28\alpha -105
$$
Hence
$$
f\left(\beta\right) = \beta^{3} - 7\beta +7 = 21\alpha^{2} +28\alpha -105 -7\left(3\alpha^{2}+4\alpha-14\right)+7 = 21\alpha^{2} +28\alpha -105 - 21\alpha^{2} -28\alpha +98 + 7 = 0.
$$
I don't know where exactly is your mistake but it must be at the computation of $\beta^{3}$, also $\beta \neq \alpha$, because if we had $\beta = \alpha$ then:
$$
\alpha = 3\alpha^{2}+4\alpha-14
$$
$$
\alpha^{2} = -5\alpha^{2}-7\alpha+28
$$
but these two equations have different roots, which is a contradiction.
