Question

find the equation whose roots are the squares of the roots of the equation x³+2x²-x+5=0

Answer 1

x^3+ax^2+bx+c= (x−α)(x−β)(x−γ) = =x3−(α+β+γ)x^2+(αβ+βγ+γα)x−αβγ


α+β+γ=−a = -2
αβ+βγ+γα=b = -1
αβγ=−c = -5

(x−α^2)(x−β^2)(x−γ^2)

=x3−(α^2+β^2+γ^2)x^2+(α^2β^2+β^2γ^2+γ^2α^2)x−α^2β^2γ^2


(α^2+β^2+γ^2) = (α+β+γ)-2(αβ+βγ+γα) = -2+2 = 0
(α^2β^2+β^2γ^2+γ^2α^2) =
α^2β^2γ^2 = (-5)^2 = 25