Question

let alpha and beta be the zeros of the cubic polynomial x^3 + ax^2 + bx + c satisfying the equation alpha*beta + 1 = 0. prove that c^2 + ac + b +1 = 0

Answer 1

$\alpha\beta+1=0\implies \alpha\beta=-1$

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

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

$\alpha\beta+\beta\gamma+\gamma\alpha=b\\-1+\gamma(\beta+\alpha)=b\\-1+c(-c-a)=b\\c^2+ac+b+1=0$

Answer 2

  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