Finding a cubic equation using Vieta's Formula

[Cqlc]

Vieta's formula states that:
In a cubic equation x^3+ax^2+bx+c=0 with roots p q and r
p+q+r=-a
pq+qr+pr=b
pqr=-c

Thus, given equation x^3-6x-3=0 with roots x1 x2 and x3, find a cubic equation with integer coefficients solutions 1/x1-1, 1/x2-1 and 1/x3-1

 

[Dr.Peterson]

Vieta's formula states that:
In a cubic equation x^3+ax^2+bx+c=0 with roots p q and r
p+q+r=-a
pq+qr+pr=b
pqr=-c

Thus, given equation x^3-6x-3=0 with roots x1 x2 and x3, find a cubic equation with integer coefficients solutions 1/x1-1, 1/x2-1 and 1/x3-1

Please show what you have tried, so we can see where you need help.

Also, confirm whether 1/x1-1 means what it says (namely, [imath]\frac{1}{x_1}-1[/imath]), or was intended to be 1/(x1-1) (that is, [imath]\frac{1}{x_1-1}[/imath]).

 

[Steven G]

If ax^3 + bx^2 + cx + d = 0 has roots x=p, x=q and x=r, then the roots for dx^3 + cx^2 + bx + a = 0 will be x= 1/p, x= 1/q and x= 1/r.
Continue from here.

 

[Cqlc]

solved thanks