Answers

2015-04-14T19:34:07+05:30
\alpha\beta+1=0\implies \alpha\beta=-1

let \gamma be the third root of given cubic :
\alpha\beta\gamma = -c \implies (-1)\gamma=-c\implies \gamma = c

\alpha+\beta+\gamma = -a \implies \alpha+\beta=-c-a

\alpha\beta+\beta\gamma+\gamma\alpha=b\\-1+\gamma(\beta+\alpha)=b\\-1+c(-c-a)=b\\c^2+ac+b+1=0
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i think u r quite good at maths
haha not really but ty :)
by the way in which class are u
whcih classes you people are in
class 10
2015-04-15T03:26:52+05:30

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  given    αβ + 1 = 0   =>      αβ = - 1      -- (1)

α , β,  δ are the roots of the polynomial:  then it is equal to the product:

   (x  - α ) (x - β)  (X - δ)
        =  x³ -  (α+β+δ)  x²  + ( αβ + βδ + δα) x - αβδ

so  we have
     αβ δ =  -  c
                 by (1)      =>  c = δ            --- (2)
      α + β + δ =  -  a
                    α + β  = - (a + δ)            ---- (3)
      αβ + βδ + δα  =  b
             =>    -1 + βδ + δα  =  b
             =>      βδ + δα = b + 1       --- (4)
          

Now to prove 

   c² + ac + b + 1
              
=  δ² + a δ + βδ + δα          using the above  (2)  , (4)
               =  δ (δ + a + β + α)       
               =  δ (δ + a - a - δ )              using  (3)
               =  δ ( 0 )
         = 0


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