Let `p,q,r` be roots of cubic `x^(3)+2X^(2)+3x+3=0`, then

Options

A. `(p)/(p+1) + (q)/(q+1) + (r )/(r+1) = 5`
B. `((p)/(p+1))^(3) + ((q)/(q+1))^(3) + ((r )/(r+1))^(3) = 45`
C. `((p)/(p+1))+((q)/(q+1))+((r )/(r+1)) = 6`
D. `((p)/(p+1))^(3)+((q)/(q+1))^(3)+((r )/(r+1))^(3)= 46`

Answer

A

Explanation

Using transformation, let
`((p)/(p+1)) = y implies p= ((y)/(1-y))`
As,p is root of cubic so
`y^(3)-5y^(2)+6y-3=0`
Now, `y_(1), y_(2),y_(3)` are roots of above equation so
` ((p)/(p+1)+(q)/(q+1)+(r )/(r+1))=5`
Also `Sigmay_(1)=5,Sigmay_(1)y_(2)=6, Sigmay_(1)y_(2)y_(3)=3`
`therefore y_(1)^(3)+y_(2)^(3)+y_(3)^(3)-3y_(1)y _(2)y_(3)`
`= Sigmay_(1)((Sigmay_(1))^(2)-3SigmaY_(1)y_(2))`
`implies y_(1)^(3)+y_(2)^(3)+y_(3)^(3)-3(3)=5((5)^(2)-3(6))`
`impliesy_(1)^(3)+y_(2)^(3)+y_(3)^(3)=44`
` therefore ((p)/(p+1))^(3)+((q)/(q+1))^(3)+((r)/(r+1))^(3)=44`