What's $\alpha+\beta$ if we have $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$? Here $\alpha$ and $\beta$ are real.
Firstly, I subtracted the two equations and got the following:
$$\alpha^3-\beta^3-6(\alpha^2-\beta^2)+13(\alpha-\beta)=-18$$ Then I tried to factorize the left hand side as: $$(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2-6\alpha-6\beta+13)=-18$$
At this point it seems we can't go on!
Then I tried adding the two equations as: $$(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2-6\alpha-6\beta+13)+12\alpha\beta=20$$
unfortunately, again I can't continue!
Is there any special creativity needed for solving this question?