Finding polynomial with roots $1/\alpha$, $1/\beta$, $1/\gamma$ for the roots of $x^3-4x^2+x+6=0$ If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3-4x^2+x+6=0$, find the equation whose roots are
1) $1/\alpha$, $1/\beta$, $1/\gamma$
2) $\alpha^2$, $\beta^2$, $\gamma^2$
How to do these kind of questions. I am stuck.
I know, 
$\alpha+\beta+\gamma=-b/a$
$\alpha\beta+\beta \gamma+\alpha \gamma=c/a$
$\alpha \beta\gamma=-d/a$

For the first one, set $x=1/y\implies$  $$\left(\dfrac1y\right)^3-4\left(\dfrac1y\right)^2+\left(\dfrac1y\right)+6=0$$
Multiply by $y^3$
For the second, $x^2=y,$
$$x(x^2+1)=4x^2-6\implies x(y+1)=4y-6$$
Square both sides and replace $x^2$ with $y$
