<s>[INST] 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.
 [/INST] Adding both the equations we have $$\alpha^3-6\alpha^2+13\alpha+\beta^3-6\beta^2+13\beta=20$$
Which is equivalent to 
$$(\alpha-2)^3+(\beta-2)^3+(\alpha-2)+(\beta-2)=0$$
Let $\alpha-2=a$ and $\beta -2=b$
Now we've ;
$$a^3+b^3+a+b=0$$
$$(a+b)(a^2-ab+b^2+1)=0$$
This give us 
$$a+b=0 \implies \color{blue}{\alpha+\beta=4}$$
Looking the given equation, we  can conclude that $\alpha,\beta >0 $ and thus another factor cannot be zero.
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