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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 $\alpha+\beta$

  1. Are the three roots of both cubic equations unique, or is there only one root? How can you prove it?
  2. 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?

Answers