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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

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by agrawal

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by agrawal

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let be the third root of given cubic :

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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

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) =>

α + β + δ = - a

αβ + βδ + δα = b

=> -1 + βδ + δα = b

=>

Now to prove

= δ (δ + a + β + α)

= δ (δ + a - a - δ ) using (3)

= δ ( 0 )

= 0