x⁴ - x - 2 = 0

x⁴ + (x³ - x³) + (x² - x²) - (2x - x) - 2 = 0

x⁴ + x³ - x³ + x² - x² - 2x + x - 2 = 0

x⁴ - x³ + x² - 2x + x³ - x² + x - 2 = 0

(x⁴ - x³ + x² - 2x) + (x³ - x² + x - 2) = 0

x.(x³ - x² + x - 2) + (x³ - x² + x - 2) = 0

(x + 1).(x³ - x² + x - 2) = 0

First case: (x + 1) = 0 → x + 1 = 0 → x = - 1

Second case: (x³ - x² + x - 2) = 0

x³ - x² + x - 2 = 0 → it's necessary to eliminate the term at the power 2

x³ - x² + x - 2 = 0 → let: x = z + (1/3)

[z + (1/3)]³ - [z + (1/3)]² + [z + (1/3)] - 2 = 0

{[z + (1/3)]².[z + (1/3)]} - [z² + (2/3).z + (1/9)] + z + (1/3) - 2 = 0

{[z² + (2/3).z + (1/9)].[z + (1/3)]} - z² - (2/3).z - (1/9) + z + (1/3) - 2 = 0

z³ + (1/3).z² + (2/3).z² + (2/9).z + (1/9).z + (1/27) - z² - (2/3).z - (1/9) + z + (1/3) - 2 = 0

z³ + (2/3).z - (47/27) = 0 ← no more term with power 2

z³ + (2/3).z - (47/27) = 0 → let: z = u + v

(u + v)³ + (2/3).(u + v) - (47/27) = 0

[(u + v)².(u + v)] + (2/3).(u + v) - (47/27) = 0

[(u² + 2uv + v²).(u + v)] + (2/3).(u + v) - (47/27) = 0

[u³ + u²v + 2u²v + 2uv² + uv² + v³] + (2/3).(u + v) - (47/27) = 0

[u³ + v³ + 3u²v + 3uv²] + (2/3).(u + v) - (47/27) = 0

[(u³ + v³) + (3u²v + 3uv²)] + (2/3).(u + v) - (47/27) = 0

[(u³ + v³) + 3uv.(u + v)] + (2/3).(u + v) - (47/27) = 0

(u³ + v³) + 3uv.(u + v) + (2/3).(u + v) - (47/27) = 0 → you can factorize: (u + v)

(u³ + v³) + (u + v).[3uv + (2/3)] - (47/27) = 0 → suppose that: [3uv + (2/3)] = 0 ← equation (1)

(u³ + v³) + (u + v).[0] - (47/27) = 0

(u³ + v³) - (47/27) = 0 ← equation (2)

You can get a system of 2 equations:

(1) : [3uv + (2/3)] = 0

(1) : 3uv = - 2/3

(1) : uv = - 2/9

(1) : u³v³ = - 8/729

(2) : (u³ + v³) - (47/27) = 0

(2) : u³ + v³ = 47/27

Let: U = u³

Let: V = v³

You can get a new system of 2 equations:

(1) : UV = - 8/729 ← this is the product P

(2) : U + V = 47/27 ← this is the sum S

You know that the values U & V are the solutions of the following equation:

x² - Sx + P = 0 ← do not confuse with the x item in you initial equation

x² - (47/27).x - (8/729) = 0

Δ = (47/27)² - [4 * - (8/729)]

Δ = 2241/27²

Δ = 83/27

x₁ = [(47/27) + (1/3).√(83/3)]/2 = (47/54) + (1/6).√(83/3) ← this is U

x₂ = [(47/27) - (1/3).√(83/3)]/2 = (47/54) - (1/6).√(83/3) ← this is V

Recall: u³ = U → u = U^(1/3) = [(47/54) + (1/6).√(83/3)]^(1/3)

Recall: v³ = V → v = V^(1/3) = [(47/54) + (1/6).√(83/3)]^(1/3)

Recall: z = u + v

Recall: x = z + (1/3)

x = (u + v) + (1/3)

x = [(47/54) + (1/6).√(83/3)]^(1/3) + [-0,184510608368363] + (1/3)

x ≈ 1.35320996419932

f(x)= x^4-x-2

f'(x) = 4x^3 - 1

x1 = 1

x2= x1 - f(x1)/f'(x1)

x2 = 1 - f(1) /f'(1)

f(1) = 1^4-1-2 = -2

f'(1) = 4(1)^3 - 1 = 3

x2 = 1 - (-2) /3

x2 = 1+2/3 = 5/3

x2 = 1.66667

guessing that this is actually x⁴ − x − 2 = 0

f = x⁴ − x − 2

f' = 4x³ – 1

x₀ = 1

x₁ = x₀ – (f(x₀) / f'(x₀))

x₁ = 1 – [(1–1–2) / (4–1)] = 1 – (–2/3) = 5/3 (this is your answer)

go one more, for curosity

x₂ = x₁ – (f(x₁) / f'(x₁))

x₂ = (5/3) – [ ((625/81) – (5/3) – 2)) / ((125•4/27) – 1) ]

x₂ = 1.666667 – [ (7.716049 – 1.666667 – 2) / (17.51852) ]

x₂ = 1.666667 – [ (4.049382) / (17.51852) ]

x₂ = 1.666667 – 0.2311487

x₂ = 1.435518