Can you Solve Oxford University Pure Mathematics Entrance Exam ?

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√{3√[3.(x + 1).√{3.(x + 1)}]} = 9 → where: (x + 1) ≥ 0 → x ≥ - 1 3√[3.(x + 1).√{3.(x + 1)}] = 81 √[3.(x + 1).√{3.(x + 1)}] = 27 3.(x + 1).√{3.(x + 1)} = 729 (x + 1).√{3.(x + 1)} = 243 √{3.(x + 1)} = 243/(x + 1) 3.(x + 1) = 243²/(x + 1)² 3 = 243²/(x + 1)³ 3.(x + 1)³ = 243² (x + 1)³ = 19683 (x + 1)³ - 19683 = 0 (x + 1)³ - 27³ = 0 → recall: a³ - b³ = (a - b).(a² + ab + b²) [(x + 1) - 27].[(x + 1)² + 27.(x + 1) + 27²] = 0 (x + 1 - 27).(x² + 2x + 1 + 27x + 27 + 729) = 0 (x - 26).(x² + 29x + 757) = 0 First case: (x - 26) = 0 x - 26 = 0 x = 26 Second case: (x² + 29x + 757) = 0 x² + 29x + 757 = 0 Δ = 29² - (4 * 757) = 841 - 3028 = - 2187 ← negative value, we stop here