<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] Lets create a polynomial for both these equations. 
Let $A(x) = x^3 - 6x^2 +13x -1 $. and $B(x) = A(x) -18$. Both these polynomials have only one real root, which is easily verified from their derivative.
Now let $\alpha+\beta = a$. Then $\beta = a-\alpha$. So transforming $A(x)$ such that $x \to a-x$ should give a polynomial whose root is $\beta$, that is  $pB(x)$, here we get $p = -1$ by comparing coefficient of $x^3$.
$$A(a-x) = p B(x) \\
(a-x)^3 - 6(a-x)^2 + 13(a-x) - 1 = -(x^3 -6x^2 +13x -19)$$
These polynomials are identically equal, so we can simply compare coefficients of equal powers of $x$. 


*

*Comparing coefficient of $x^3$ we get $p=-1$.

*Comparing coefficient of $x^2$ : $3a -6 = 6$, we get $a = 4$

*Comparing coefficient of $x$: $-3a^2+12a-13 = -13$. Now, $a = 4$ satisfies this equation.

*Comparing constant term: $a^3-6a^2+13a-1 = 19$. Again $a = 4$ satisfies this.
We see $a = 4$ is the only value which satisfies all three conditions.
Hence $\alpha + \beta = a = 4$
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