Complex root of equation 1 let $a,b$ & $c$ are the roots of cubic $$x^3-3x^2+1=0$$
Find a cubic whose roots are $\frac a{a-2},\frac b{b-2}$  and $\frac c{c-2} $
hence or otherwise find value of $(a-2)(b-2)(c-2)$
 A: (see figure below)
Invert the given relationship $$y=\dfrac{r}{r-2} \ \ \ \ (1)$$ into $$r=\dfrac{2y}{y-1} \ \ \ \ (2)$$
As $r$ is a root of the given cubic, we have $$r^3-3r^2+1=0 \ \ \ \ (3)$$
It suffices then to plug relationship (2) into (3) to get on the LHS a rational expression ; equating its numerator to zero gives : 
$$3y^3 - 9y^2 - 3y + 1=0 \ \ \ \ (4)$$
The following picture gives a graphical explanation of the correspondance of the two equations (3) and (4), or more exactly the curves of $f$ and $g$ defined by $f(r)=r^3-3r^2+1$ (red curve) and $g(y)=-3y^3+9y^2+3y-1$ (blue curve, reversed) and their roots that ares omewhat ''mirrored'' the ones into the others by the (black) transformation curve defined by $y:=\varphi(r)=r/(r-2)$.

The Matlab program that has generated the above figure :
clear all;close all;hold on
LW='linewidth';a=-3;b=4;
text(b-0.3,0.2,'r');text(0.2,b-0.3,'y');
axis([a,b,a,b]);plot([a,b],[0,0],'k');plot([0,0],[a,b],'k');
r=a:0.01:b; y=a:0.01:b;
phi=@(r)(r./(r-2));
plot(r,phi(r),'k',LW,2);
plot(r,r.^3-3*r.^2+1,'r',LW,1)
plot(-3*y.^3+9*y.^2+3*y-1,y,LW,1)
R=roots([1,-3,0,1]);% roots of 1r^3-3r^2+0r+1=0
Y=f(R); % same as roots([-3,9,3,-1]) i.e., roots of -3y^3+9y^2+3*y-1=0
for k=1:3
   plot([R(k),R(k),0],[0,Y(k),Y(k)],'ko-')
end;

A: HINT
We have that $(x-a)(x-b)(x-c)=x^3-3x^2+1$ and then


*

*$a+b+c=3$

*$ab+bc+ca=0$

*$abc=-1$
A: You know that: $$a+b+c=3$$ $$ab+bc+ac=0$$ $$abc=-1$$
Also, $$(a-2)(b-2)(c-2)=abc-2(ab+bc+ac)+4(a+b+c)-8$$

So $$(a-2)(b-2)(c-2)=-1-2(0)+4(3)-8=3$$

For cubic with roots as $$\frac a{a-2},\frac b{b-2},\frac c{c-2}$$
Find out $$\frac a {a-2}+\frac b{b-2}+\frac c{c-2}$$
and $$\frac a{a-2}\cdot \frac b{b-2} + \frac b{b-2}\cdot \frac c{c-2} + \frac a{a-2}\cdot \frac c{c-2}$$ by taking LCM and rigorous solving.  
You already know that $$\frac a{a-2}\cdot \frac b{b-2} \cdot \frac c{c-2}=3$$
