2 Useful Methods for a Tricky Radical Equation!

[@gregevgeni1864]

Obviously x²-x-2=(x-2)(x+1)≠0 , i.e. x≠2 and x≠-1.(*) The given equation is equivalent to ³√(((x²-x+1)(x+1))/(x-2)(x+1)) - ³√(((x²+2x+2²)(x-2))/(x+1)(x-2))= =³√(27/(x+1)(x-2)) ³√((x³+1)/(x-2)(x+1))- -³√((x³-8)/(x+1)(x-2))- -³√((27/(x+1)(x-2))= 0 (1). Let A=³√((x³+1)/(x-2)(x+1)²), B=³√((x³-8)/(x+1)(x-2)), C=³√((27/(x+1)(x-2)) (2) Because (1) => A+(-B)+(-C)=0 => A³-B³-C³=3•A•B•C (**) known identity. Due to (**) and (2) we have [³√((x³+1)/(x-2)(x+1)²)]³- -[³√((x³-8)/(x-2)²(x+1))]³- -[³√((27/(x-2)(x+1)]³= = 3•³√((27•(x³+1)(x³-8))/(x-2)³(x+1)³)) => (x³+1)/(x²-x-2) -(x³-8)/(x²-x-2) - -27/(x²-x-2) = =3•³√(3³•(x³+1)(x³-8)/(x²-x-2)³) => -18/(x²-x-2)=3•3(³√(x³+1)(x³-8))/(x²-x-2) => -2 =³√(x³+1)(x³-8), because (*), (-2)³=(x³+1)(x³-8)=>-8=x⁶-7x³-8 => x⁶-7x³=0 => x³(x³-7)=0 => x=0 or x=³√7 .

[@Fjfurufjdfjd]

Πρεπει χ=/2 χ=/-1 (χ^2-χ+1)^(1/3)/(χ-2)^(1/3)-(χ^2+2χ+4)^(1/3)/(χ+1)^(1/3)=3/[(χ-2)(χ+1)]^(1/3) Πολλαπλασιαζω με [(χ-2)(χ+1)]^(1/3) ...[χ^3+1^3]^(1/3)-[χ^3-2^3]^(1/3)=3 Θετω χ^3+1=α^3 και χ^3-8=β^3 α-β=3 α^3-β^3=9 ....α-β=3 αβ=-2 α=1 , β=-2 ή α=2 , β=-1 χ=0 (τριπλη) ή χ=(7)^(1/3) (τριπλη) δεκτες.

[@Quest3669]

X= 7^1/3; 0

[@walterwen2975]

A Tricky Radical Equation: ³√[(x² - x + 1)/(x - 2)] - ³√[(x² + 2x + 4)/(x + 1)] = ³√[27/(x² - x - 2)], x ϵ R; x =? x ϵ R, x - 2 ≠ 0, x + 1 ≠ 0 ³√[(x² - x + 1)/(x - 2)] - ³√[(x² + 2x + 4)/(x + 1)] = {³√[(x + 1)(x² - x + 1)] - ³√[(x - 2)(x² + 2x + 4)]}/{³√[(x - 2)(x + 1)]} ³√[27/(x² - x - 2)] = 3/{³√[(x - 2)(x + 1)]} ³√[(x + 1)(x² - x + 1)] - ³√[(x - 2)(x² + 2x + 4)] = ³√(x³ + 1) - ³√(x³ - 2³) = 3 Let: u = ³√(x³ + 1), v = ³√(x³ - 2³); u - v = 3, u³ - v³ = (x³ + 1) - (x³ - 2³) = 9 u³ - v³ = (u - v)(u² + uv + v²) = (u - v)[(u - v)² + 3uv] = 3(9 + 3uv) = 9(3 + uv) = 9 3 + uv = 1, uv = - 2; u - v = 3, v = u - 3, uv = u(u - 3) = - 2, u² - 3u + 2 = 0 (u - 1)(u - 2) = 0, u - 1 = 0, u = 1; uv = - 2, v = - 2 or u - 2 = 0, u = 2, v = - 1 u = ³√(x³ + 1) = 1, x³ + 1 = 1, x = 0 or u = ³√(x³ + 1) = 2, x³ = 7, x = ³√7 v = ³√(x³ - 2³) = - 2, x³ - 2³ = - 2³, x = 0 or v = ³√(x³ - 2³) = - 1, x³ = 7, x = ³√7 Answer check: ³√[(x + 1)(x² - x + 1)] - ³√[(x - 2)(x² + 2x + 4)] = ³√(x³ + 1) - ³√(x³ - 2³) = 3 x = 0: ³√(x³ + 1) - ³√(x³ - 2³) = 1 - ³√(- 2³) = 1 - (- 2) = 3; Confirmed x = ³√7: ³√(7 + 1) - ³√(7 - 8) = 2 - (- 1) = 3; Confirmed Final answer: x = 0 or x = ³√7: