Question:
The equation $x^3 -x^2 - 3x - 10 = 0$ has roots $\alpha,\,\beta,\,\gamma$
Let $u = -\alpha + \beta + \gamma$. Show that $u + 2\alpha = 1$, and hence find a cubic equation having roots $\alpha + \beta + \gamma,\;\alpha - \beta + \gamma,\; \alpha + \beta - \gamma $
Try:
$$\alpha +\beta +\gamma =1\\\alpha \beta +\alpha \gamma +\beta \gamma = -3\\\alpha \beta \gamma =10$$ $$u+2\alpha=-\alpha+\beta+\gamma+2\alpha\\u+2\alpha=\alpha +\beta +\gamma\\u+2\alpha=1$$ I can not figure out what to do next.