If α, β, γ are the roots of the equation x³ - px² + qx + r = 0 then find the value of (α + β)(β + γ)(γ + α)
Answers
EXPLANATION.
α, β, γ are the roots of the equation.
⇒ x³ - px² + qx + r = 0.
As we know that,
Sum of the zeroes of the cubic polynomial.
⇒ α + β + γ = - b/a.
⇒ α + β + γ = -(-p)/1 = p.
Sum of the zeroes of the cubic polynomial two at a time.
⇒ αβ + βγ + γα = c/a.
⇒ αβ + βγ + γα = q/1 = q.
Products of the zeroes of the cubic polynomial.
⇒ αβγ = - d/a.
⇒ αβγ = -(r/1) = - r.
To find value of :
⇒ (α + β)(β + γ)(γ + α).
As we know that,
We can write equation as,
⇒ [αβ + αγ + β² + βγ] x (γ + α).
⇒ γ(αβ + αγ + β² + βγ) + α(αβ + αγ + β² + βγ).
⇒ (αβγ + αγ² + β²γ + βγ²) + (α²β + α²γ + αβ² + αβγ).
⇒ αγ² + β²γ + βγ² + α²β + α²γ + αβ² + αβγ + αβγ.
⇒ (α²β + α²γ + αβγ) + (β²γ + β²α + αβγ) + (γ²α + γ²β + αβγ) - αβγ.
⇒ α(αβ + αγ + βγ) + β(βγ + βα + αγ) + γ(γα + γβ + αβ) - αβγ.
⇒ (α + β + γ)(αβ + βγ + γα) - αβγ.
Put the values in the equation, we get.
⇒ (p)(q) - (-r).
⇒ pq + r.
Values of : (α + β)(β + γ)(γ + α) = pq + r.
Answer:
pq + r
Step-by-step explanation:
Given
α, β, γ are the roots of the equation x³ - px² + qx + r = 0
Comparing the given equation with ax³ + bx² + cx + d = 0
- a = 1
- b = - p
- c = q
- d = r
Sum of roots = - b/a
⇒ α + β + γ = - ( - p) /1
⇒α + β + γ = p
To solve the question we will write the above equation as
- α + β = p - γ
- β + γ = p - α
- γ + α = p - β
Sum of product of roots taken two at a time = c/a
⇒ αβ + βγ + γα = q/1 = q
Product of roots = - d/a
⇒ αβγ = - r/1
⇒ αβγ = - r
Now, coming to the question
( α + β )( β + γ )( γ + α )
= ( p - γ )( p - α)( p - β )
= p³ - ( α + β + γ )p² + ( αβ + βγ + γα )p - αβγ
= p³ - ( p)p² + ( q)p - ( - r)
= p³ - p³ + pq + r
= pq + r
Therefore the value of the given expression is pq + r.