If α,β,γ are the roots of x3+2x2−3x−1=0, then α−2+β−2+γ−2= Explain please?

If the original equation is x^3 + 2x^2 - 3x - 1 = 0, the sum of the roots is -2, therefore:
α + β + γ = -2
α − 2 + β − 2 + γ − 2 = α + β + γ - 2 - 2 - 2 = -2 - 2 - 2 - 2 = -8

α, β, γ are the roots of x³ + 2x² - 3x - 1 = 0
α + β + γ = -2
αβ + βγ + αγ = -3
αβγ = 1

αβ + βγ + αγ = -3
(αβ + βγ + αγ)² = (-3)²
α²β² + β²γ² + α²γ² + 2α²βγ + 2αβ²γ + 2αβγ² = 9
(α²β² + β²γ² + α²γ²) + 2αβγ(α + β + γ) = 9
(α²β² + β²γ² + α²γ²) + 2*1*(-2) = 9
(α²β² + β²γ² + α²γ²) - 4 = 9
α²β² + β²γ² + α²γ² = 13

α⁻² + β⁻² + γ⁻²
= (1/α²) + (1/β²) + (1/γ²)
= (β²γ²/α²β²γ²) + (α²γ²/α²β²γ²) + (α²β²/α²β²γ²)
= (α²β² + β²γ² + α²γ²) / (αβγ)²
= 13/(1)²
= 13

i) What do you require? Is it α−2+β−2+γ−2 or α⁻² + β⁻² + γ⁻²?

ii) Applying properties of roots of third degree polynomials,
Sum of roots = α + β + γ = -b/a = -2/1 = -2 -------- (1)
Sum of product of roots taken two at a time = αβ + βγ + αγ = c/a = -3/1 = -3 ------- (2)
Product of roots = -d/a = 1/1 = 1 ------ (3)

iii) If required is: α−2+β−2+γ−2
Then it is = α + β + γ - 6 = -2 - 6 = -8 [Substituting the value α + β + γ = -2 from (1)]

iv) Considering the requirement is: α⁻² + β⁻² + γ⁻², the solution is as below:

1/α + 1/β + 1/γ = (αβ + βγ + αγ)/αβγ = -3 [Substituting the respective values from (2) & (3)]

Squaring this, 1/α² + 1/β² + 1/γ² + 2(1/αβ + 1/βγ + 1/αγ) = 9
==> (1/α² + 1/β² + 1/γ²) + 2{(α + β + γ)/αβγ} = 9
==> (1/α² + 1/β² + 1/γ²) + 2(-2)/1 = 9
==> (1/α² + 1/β² + 1/γ²) = 13

Thus, α⁻² + β⁻² + γ⁻² = 13