A Nice Math Olympiad Radical Equation | Find the Value of x 

A Nice Math Olympiad Radical Equation: x²(√x) + (3 - √x)⁵ = 33; x ϵR; x = ? x²(√x) + (3 - √x)⁵ = (√x)⁵ + (3 - √x)⁵ = 33; 5 > x > 0 Let: u = √x, v = 3 - √x; u⁵ + v⁵ = 33, u + v = 3, uv = (√x)(3 - √x) = 3√x - x (u + v)² = u² + v² + 2uv = 3² = 9, u² + v² = 9 - 2uv (u + v)³ = u³ + v³ + 3uv(u + v) = 3³ = 27, u³ + v³ = 27 - 3uv(3) = 9(3 - uv) (u³ + v³)(u² + v²) = u⁵ + v⁵ + u²v²(u + v) = 9(3 - uv)(9 - 2uv) u⁵ + v⁵ = 9(3 - uv)(9 - 2uv) - 3u²v² = 33, 3(27 - 15uv + 2u²v²) - u²v² = 11 70 - 45uv + 5u²v² = 0, u²v² - 9uv + 14 = 0, (uv - 7)(uv - 2) = 0 uv - 7 = 0, uv = 7 or uv - 2 = 0, uv = 2 uv = 7 = 3√x - x, x - 3√x + 7 = 0, √x = (3 ± i√17)/2; x ϵR, Rejected uv = 2 = 3√x - x, x - 3√x + 2 = 0, (√x - 1)(√x - 2) = 0 √x - 1 = 0, √x = 1, x = 1 or √x - 2 = 0, √x = 2, x = 4 Answer check: x = 1: x²(√x) + (3 - √x)⁵ = 1 + 2⁵ = 1 + 32 = 33; Confirmed x = 4: 4²(√4) + (3 - √4)⁵ = 16(2) + 1⁵ = 32 + 1 = 33; Confirmed Final answer: x = 1 or x = 4