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

[@潘博宇-k4l]

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

[@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).

[@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
.....

[@icfj77]

x= -2; x=1+√5 (not x=1-√5)