Find $\alpha + \beta$ given that: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ , ( $\alpha , \beta \in \Bbb R)$ 
Find $\alpha + \beta$ given that: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ , ( $\alpha , \beta \in \Bbb R)$

I have a solution involving a variable change ($\alpha=x+2$) which is not beautiful in my opinion! I'm looking for a more beautiful solution.

No change of variables:
Adding the two equations and using $$\alpha^2 + \beta^2 = (\alpha+\beta)^2-2\alpha\beta\\
\alpha^3+\beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$gives the equation
$$(\alpha + \beta)^3 - 6(\alpha + \beta)^2 + (13-3\alpha\beta)(\alpha + \beta) + 12\alpha\beta - 20$$
meaning that $\alpha + \beta$ is the root of the polynomial $$x^3-6x^2+(13-3\alpha\beta)x + 12\alpha\beta - 20$$
This polynomial has only one real solution, $x=4$.
