How to find roots of cubic equation? The roots of equation \begin{equation*}36x^3+72x^2+23x-5=0\end{equation*} are $\alpha$, $\beta$ and $\gamma$. Find a cubic equation with numerical coefficients, which has the roots 6$\alpha-1$, 6$\beta-1$ and 6$\gamma-1$ without solving for $\alpha$, $\beta$ or $\gamma$.
I know that the sum of roots $\alpha$+$\beta$+$\gamma$ is given by -$\frac{72}{36}$ and the product of roots $\alpha$$\beta$$\gamma$ is given by -$\frac{-5}{36}$, but when I expand the equation \begin{equation*}(x-(6\alpha-1))(x-(6\beta-1))(x-(6\gamma-1))\end{equation*}I get a complicated expression with the terms 36$\alpha$$\beta$+36$\alpha$$\gamma$+36$\beta$$\gamma$. How do I carry on from there, or is there an easier way to solve this?
 A: 
I get a complicated expression

That's still a symmetric expression in $\alpha,\beta,\gamma$ so you can express it in terms of the elementary symmetric polynomials $\alpha+\beta+\gamma=-72/36\,$, $\alpha\beta+\alpha\gamma+\beta\gamma=23/36\,$, $\alpha\beta\gamma=5/36\,$.

is there an easier way to solve this

Let $\,y = 6x - 1\,$ and substitute $\,x = (y+1)/6\,$ into the original equation, then expand and collect. The resulting polynomial in $y$ will have precisely the roots $6\alpha-1, 6\beta-1, 6\gamma-1\,$.
A: It's not hard. It takes five minutes of work.
$$\alpha+\beta+\gamma=-\frac{72}{36}=-2,$$
$$\alpha\beta+\alpha\gamma+\beta\gamma=\frac{23}{36}$$ and
$$\alpha\beta\gamma=\frac{5}{36}.$$
Thus, $$\sum_{cyc}(6\alpha-1)=6(-2)-3=-15,$$
$$\sum_{cyc}(6\alpha-1)(6\beta-1)=\sum_{cyc}(36\alpha\beta-12\alpha+1)=36\cdot\frac{23}{36}-12(-2)+3=50$$ and
$$\prod_{cyc}(6\alpha-1)=216\alpha\beta\gamma-36(\alpha\beta+\alpha\gamma+\beta\gamma)+6(\alpha+\beta+\gamma)-1=$$
$$=216\cdot\frac{5}{36}-36\cdot\frac{23}{36}+6(-2)-1=-6,$$
which gives the answer:
$$x^2+15x^2+50x+6=0$$
