Can you Pass Oxford University Admission Exam ? | Find x=?

[@oahuhawaii2141]

There's a shortcut:
(x + 1)⁶ = 2⁶
x = 2*r6 - 1 , where r6 is the six 6th roots of 1
r6 = ±1, ±1/2 ± i*√3/2
2*r6 = ±2, ±1 ± i*√3
x = 2*r6 - 1 = 1, -3, ±i*√3, -2 ± i*√3

[@superacademy247]

Thanks for sharing your method and support 💕🙏🙏💡✅

[@raghvendrasingh1289]

(x+1)^3 = - 8 , (x+1)^3 = 8
Case 1
x^3+3x^2+3x+9 = 0
(x+3)(x^2+3) = 0
x = - 3 , i√3 , -i √3
Case 2
(x+1)^3 - 2^3 = 0
(x - 1)(x^2+2x+1+4+2x+2) = 0
(x - 1) (x^2+4 x+7) = 0
x = 1 , - 2+i√3 , - 2 - i√3

[@GillesF31]

Yes ... or ...
(x + 1)⁶ = 64

(x + 1)⁶ = 2⁶

(x + 1)⁶ - 2⁶ = 0

set k = x + 1

k⁶ - 2⁶ = 0

(k³)² - (2³)² = 0

(k³ - 2³)·(k³ + 2³) = 0

(k - 2)·(k² + 2k + 4)·(k + 2)·(k² - 2k + 4)

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/// cases: (k - 2) = 0 and (k + 2) = 0

root #1: k - 2 = 0 => k = 2 => x + 1 = 2 => ■ x = 1

root #2: k + 2 = 0 => k = -2 => x + 1 = -2 => ■ x = -3

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/// case: (k² + 2k + 4) = 0

k² + 2k + 4 = 0

Δ = 2² - 4·1·4 = 4 - 16 = -12

√Δ = ±i√12 = ±2i√3

• k = (-2 + 2i√3/(2·1) = -1 + i√3

• k = (-2 - 2i√3/(2·1) = -1 - i√3

root #3: k = -1 + i√3 => x + 1 = -1 + i√3 => ■ x = -2 + i√3

root #4: k = -1 - i√3 => x + 1 = -1 - i√3 => ■ x = -2 - i√3

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/// case: (k² - 2k + 4) = 0

k² - 2k + 4 = 0

Δ = (-2)² - 4·1·4 = 4 - 16 = -12

√Δ = ±i√12 = ±2i√3

• k = (-(-2) + 2i√3/(2·1) = 1 + i√3

• k = (-(-2) - 2i√3/(2·1) = 1 - i√3

root #5: k = 1 + i√3 => x + 1 = 1 + i√3 => ■ x = i√3

root #6: k = 1 - i√3 => x + 1 = 1 - i√3 => ■ x = -i√3

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/// final results:

■ root #1: x = 1
■ root #2: x = -3
■ root #3: x = -2 + i√3
■ root #4: x = -2 - i√3
■ root #5: x = i√3

■ root #6: x = -i√3
🙂

[@superacademy247]

Nice solutions. Thanks for your feedback and sharing 👏🙏😎💯 method