Heyy, thank you for asking the question
i suppose you know what to do if roots are shifted by or multiplied by 1 or 2 towards right such type of problems
now we know from the equation αβγ= -r
αβ+βγ+αγ=q
α+β+γ =0
(α+β+γ)2 = α2+β2+γ2+ 2αβ+2βγ+2αγ
0=α2+β2+γ2+2q
α2+β2+γ2= -2q
now consider new roots β/γ+γ/β=β2+γ2/βγ
= ( -2q - α2)/(-r/α)
= ( 2qα + α3)/r
but we have α3 + qα+r =0 substituting this in above equaiton we get
= (qα – r)/r
= q/r*α – 1
so from the last result we came to know that roots of the new equation are q/r times of previous roots and a shift of -1 so the final equation of new roots are
(r/q)3 (x+1)3+q* r/q*(x+1)+r=0
now lets check for a root (r/q)3 ( q/r*α – 1+1)3+q* r/q*( q/r*α – 1+1)+r=0
(r/q)3 (q/r*α )3+q* r/q*( q/r*α )+r=0
α 3+q*α+r=0 [proved]
Hope you got the solution.....
