Manipulating roots of polynomials...?

Question

I am not fully understanding how using substitutions helps in finding equations with roots related to a given equation. For example, in the following problem (extract from CIE A-Level Further Math paper):

The equation x^4 - x^3 -1 =0 has roots (alpha, beta, gamma, delta). By using the substitution y=x^3, find the exact value of (alpha^6+beta^6+gamma^6+delta^6)...

Please help...

kb · Accepted Answer

Let α, β, γ, δ be the four roots of x⁴ - x³ - 1 = 0

Let y = x³.
So, the equation becomes y^(4/3) - y - 1 = 0
==> y^(4/3) = y + 1
==> y⁴ = (y + 1)³
==> y⁴ - y³ - 3y² - 3y - 1 = 0.
Note that this equation has roots α³, β³, γ³, δ³.

So, we have y⁴ - y³ - 3y² - 3y - 1 = (y - α³) (y - β³) (y - γ³) (y - δ³).

Expanding the right side yields
y⁴ - (α³ + β³ + γ³ + δ³) y³ + (α³β³ + α³γ³ + α³δ³ + β³γ³ + β³δ³ + γ³δ³)y² - 
(α³β³γ³ + α³β³δ³ + α³γ³δ³ + β³γ³δ³) y + α³β³γ³δ³.

So comparing coefficients, we obtain 
α³ + β³ + γ³ + δ³ = 1
α³β³ + α³γ³ + α³δ³ + β³γ³ + β³δ³ + γ³δ³ = -3
α³β³γ³ + α³β³δ³ + α³γ³δ³ + β³γ³δ³ = 3
α³β³γ³δ³ = -1.
------------------
Moving onward...

Squaring both sides of α³ + β³ + γ³ + δ³ = 1 yields
(α⁶ + β⁶ + γ⁶ + δ⁶) + 2(α³β³ + α³γ³ + α³δ³ + β³γ³ + β³δ³ + γ³δ³) = 1.  
==> (α⁶ + β⁶ + γ⁶ + δ⁶) + 2(-3) = 1  
==> α⁶ + β⁶ + γ⁶ + δ⁶ = 7.

I hope this helps!

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Manipulating roots of polynomials...?

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