Finding a cubic equation from the relation between the roots

Question

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.

Answer 1

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 $$x^3-(u+v+w)x^2+(uv+vw+wu)x-uvw=0\iff x^3-x^2-13x+93=0.$$

Answer 2

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.