Are you expected to solve using Newton's Sums?

For a polynomial P(x) of degree n,

P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + aₙ₋₂xⁿ⁻² + ... + a₁x + a₀

When P(x) = 0 has roots x₁ , x₂ , x₃ , ... xₙ, define the following sums to be :

P₁ = x₁ + x₂ + x₃ + ... xₙ
P₂ = x₁² + x₂² + x₃² + ... xₙ²
P₃ = x₁³ + x₂³ + x₃³ + ... xₙ³
.
.
.
Pₙ = x₁ⁿ + x₂ⁿ + x₃ⁿ + ... xₙⁿ


From Newton sums we know that :


aₙP₁ + aₙ₋₁ = 0 ①

aₙP₂ + aₙ₋₁P₁ + 2aₙ₋₂ = 0 ②

aₙP₃ + aₙ₋₁P₂ + aₙ₋₂P₁ + 3aₙ₋₃ = 0 ③


So,

Let P(x) = x⁴ - x³ + x + 2 = 0

Define the following sums :

P₁ = ∑α = α + β + γ + δ
P₂ = ∑α² = α² + β² + γ² + δ²
P₃ = ∑α³ = α³ + β³ + γ³ + δ³


Then, from ①, we get :

∑α - 1 = 0
∑α = 1

From ②, we get :

∑α² - ∑α + 2(0) = 0
∑α² = ∑α = 1


From ③, we get :


∑α³ - ∑α² + (0)∑α + 3(1) = 0

∑α³ = ∑α² - 3

= 1 - 3

= -2

Or , Using Vieta's formulas,



Let P(x) = x⁴ - x³ + x + 2 = 0. The roots are α,β,γ,δ



So (x - α)(x - β)(x - γ)(x - δ) = 0

Expanding this gives :


x⁴ - (α + β + γ + δ)x³ + (αβ + αγ + αδ + βγ + βδ + γδ)x² - (αβγ + αβδ + αγδ + βγδ)x + αβγδ = 0


Comparing coefficients,

∑α = α + β + γ + δ = 1


∑α² = (α + β + γ + δ)² - 2(αβ + αγ + αδ + βγ + βδ + γδ)

= (∑α)² - 2(0)

= 1 - 2(0)

= 1


∑1/α = 1/α + 1/β + 1/γ + 1/δ

= (βγδ + αγδ + αβδ + αβγ) / (αβγδ)

= (-1) / 2

= -½

Lastly,


Since x⁴ - x³ + x + 2 = 0,

x⁴ = x³ - x - 2

Divide throughout by x,


x³ = x² - 1 - 2/x
x³ = x² - 1 - 2(1/x)


Which means :

α³ = α² - 1 - 2(1/α)
β³ = β² - 1 - 2(1/β)
γ³ = γ² - 1 - 2(1/γ)
δ³ = δ² - 1 - 2(1/δ)


So,

∑α³ = ∑(α² - 1 - 2(1/α))

= ∑α² - ∑1 - ∑2(1/α)

= 1 - 4(1) - 2(-½)

= 1 - 4 + 1

= -2