Finding a cubic equation from the relation between the roots I'm trying to solve this problem:
$ x^3 - x^2 - 3x -10 = 0$ has roots α,β,γ.
Let u = −α+β+γ. Show that u+2α=1, and hence find a cubic equation having roots −α+β+γ, α−β+γ, α+β−γ.
I was able to complete the first part easily using Vieta's theorem, although I am having trouble with the second part.
 A: Since $$\alpha+\beta+\gamma=1,\ \ \alpha\beta+\beta\gamma+\gamma\alpha=-3,\ \ \alpha\beta\gamma=10,$$
 setting $$u=-\alpha +\beta+\gamma,\ v=\alpha-\beta+\gamma,\ w=\alpha+\beta-\gamma$$ gives you
$$u=1-2\alpha,\ v=1-2\beta,\ w=1-2\gamma.$$
So, we have 
$$u+v+w=3-2(\alpha+\beta+\gamma)=3-2\cdot 1=1$$
$$\begin{align}uv+vw+wu&=(1-2\alpha)(1-2\beta)+(1-2\beta)(1-2\gamma)+(1-2\gamma)(1-2\alpha)\\&=3-4(\alpha+\beta+\gamma)+4(\alpha\beta+\beta\gamma+\gamma\alpha)\\&=3-4\cdot 1+4\cdot (-3)\\&=-13\end{align}$$
$$\begin{align}uvw&=(1-2\alpha)(1-2\beta)(1-2\gamma)\\&=1-2(\alpha+\beta+\gamma)+4(\alpha\beta+\beta\gamma+\gamma\alpha)-8\alpha\beta\gamma\\&=1-2\cdot 1+4\cdot (-3)-8\cdot 10\\&=-93\end{align}$$
Hence, a cubic equation having roots $-\alpha+\beta+\gamma,\ \alpha-\beta+\gamma,\ \alpha+\beta-\gamma$ is
$$x^3-(u+v+w)x^2+(uv+vw+wu)x-uvw=0\iff x^3-x^2-13x+93=0.$$
A: For the second part, let $y=\alpha + \beta + \gamma -2x$, where x is a root of the original polynomial.
Then $y=1-2x \implies x=\frac{1}{2}-\frac{y}{2}$. 
Substitute this into $x^3-x^2-3x-10=0$ to find the required equation. 
