Can You Solve This Radical Equation CHALLENGE? | Algebra

[@Quest3669]

Eqn can turns to x^3-8x-8= 0 Or (x+2)(x^2-2x-4)= 0 or x= -2; +-√5+(1)solns

[@SidneiMV]

y = 2∛(x + 1) x = 2∛(y + 1) y³ = 8x + 8 x³ = 8y + 8 x³ - y³ = -8(x - y) (x - y)(x² + xy + y² + 8) = 0 x = y ∨ x² + xy + y² + 8 = 0 [ 1 ] x = y x = 2∛(x + 1) x³ = 8x + 8 x³ - 8x - 8 = 0 x³ + 8 -8(x + 2) = 0 (x + 2)(x² - 2x - 4) = 0 x + 2 = 0 ∨ x² - 2x - 4 = 0 [ 1.1 ] x + 2 = 0 *x = -2* [ 1.2 ] x² - 2x - 4 = 0 *x = 1 ± √5* [ 2 ] x² + xy + y² + 8 = 0 y³ = 8x + 8 => y⁴ = 8xy + 8y x³ = 8y + 8 => x⁴ = 8xy + 8x x⁴ + y⁴ = 16xy + 8(x + y) x² + y² = -(xy + 8) x⁴ + y⁴ + 2(xy)² = (xy)² + 64 + 16xy .....

[@潘博宇-k4l]

X=[-2, 1+(5)^(1/2),, 1-(5)^(1/2)]

[@icfj77]

x= -2; x=1+√5 (not x=1-√5)

[@Fjfurufjdfjd]

χ=-2 ή χ=1+(5)^(1/2) ή χ=1-(5)^(1/2) Θετω χ+1=ψ^3. Προσθετω και στα δυο μελη της σχεσης που δινεται το 1 και εχω ψ^3=2(1+2ψ)^(1/3)+1. Θετω 1+2ψ=α^3 Τοτε ψ^3=2α+1.δηλαδη εχω το συστημα: ψ^3=2α+1 , α^3=2ψ+1 αφαιρω κατα μελη ψ^3-α^3=2(α-ψ) (ψ-α)( ψ^2+α^2+αψ+2)=0 Αλλα ψ2+α^2+αψ=-2 2ψ^2+2α^2+ 2αψ=-4 (ψ+α)^2+α^2+ψ^2=-4 ατοπο Αρα ψ-α=0 ψ=α.τοτεεχω ψ^3-2ψ-1=0 (ψ+1)(ψ^2-ψ-1)=0 ψ=-1 ψ=[1+(5)^(1/2)]/2 ψ=[1-(5)^(1/2)]/2 Οποτε απο τη σχεση χ=ψ^3-1 εχω τελικα χ=-2 , χ=1+(5)^(1/2) , χ=1-(5)^(1/2).