The real numbers $\alpha,\beta$ satisfy
$$\alpha^3-3\alpha^2+5\alpha-17=0\tag{1}$$ $$\beta^3-3\beta^2+5\beta+11=0\tag{2}$$
Find $\alpha+\beta$
- Are the three roots of both cubic equations unique, or is there only one root? How can you prove it?
- What's the best approach to this problem?
I tried using Vieta's formulas, where the sum of three roots of (1) and (2) are:
$$\alpha_1+\alpha_2+\alpha_3=3$$ $$\beta_1+\beta_2+\beta_3=3$$
Summing both,
$$\alpha_1+\alpha_2+\alpha_3+\beta_1+\beta_2+\beta_3=6$$
Assuming there is only one root for each of (1) and (2), we are done, but what if there isn't?