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Let `p,q,r` be roots of cubic `x^(3)+2X^(2)+3x+3=0`, thenA. `(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` |
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Answer» Correct Answer - A 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` |
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