What's $\alpha+\beta$ if we have: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ ($\alpha$ and $\beta$ are Real)

Let $f(x)=x^3-6x^2+13x,$ Then $f'(x)=3x^2-12x+13 = 3[(x-2)^2]+1>0$

So $f(x)$ is strictly increasing function.

Now above we have given $f(\alpha)=1$ and $f(\beta) = 19\;,$ Where $\alpha<\beta$

Then $f(\alpha)<f(\beta)$, because the function is increasing.

Now \begin{align*}f(4-x)&=(4-x)^3-6(4-x)^2+13(4-x)\\&=64-x^3-48x+12x^2-96-6x^2+48x+52-13x\\&=-x^3+6x^2-13x+20=-f(x)+20\end{align*} So if $x=\alpha$, Then $f(4-\alpha)=-f(\alpha)+20=19=f(\beta)$

The function is continuous and monotonically increasing, thus $f(x)=f(y)\implies x=y$, hence $$4-\alpha=\beta\Rightarrow \boxed{\alpha+\beta=4}$$

In order to get rid of the quadratic terms we write $\alpha:=a+2$, $\beta:=b+2$. Then $$1=\alpha^3-6\alpha^2+13\alpha=(a+2)^3-6(a+2)^2+13(a+2)=a^3+a+10\ ,$$ and similarly $$19=\beta^3-6\beta^2+13\beta=b^3+b+10\ .$$ It follows that $a$ and $b$ satisfy the equations $$a^3+a=-9,\quad b^3+b=9\ .$$ Now (an unexpected) symmetry comes to our rescue: As $t\mapsto t^3+t$ is monotonically increasing there is exactly one $b=:b_*>0$ satisfying the second of these equations, and it is then obvious that $a:=-b_*$ is the unique solution of the first of these equations. It follows that necessarily $$\alpha+\beta=-b_*+2 +b_*+2=4\ .$$