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`        If alpha,beta,gamma be the roots of x^3+qx+r=0,then find the equation whose roots are Beta/gamma+gamma/beta,gamma/alpha+alpha/gamma,alpha/beta+beta /alpha`
one year ago

```							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=0now 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.....
```
one year ago
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