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√{3√[3.(x + 1).√{3.(x + 1)}]} = 9 → where: (x + 1) ≥ 0 → x ≥ - 13√[3.(x + 1).√{3.(x + 1)}] = 81√[3.(x + 1).√{3.(x + 1)}] = 273.(x + 1).√{3.(x + 1)} = 729(x + 1).√{3.(x + 1)} = 243√{3.(x + 1)} = 243/(x + 1)3.(x + 1) = 243²/(x + 1)²3 = 243²/(x + 1)³3.(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) = 0First case: (x - 26) = 0x - 26 = 0x = 26Second case: (x² + 29x + 757) = 0x² + 29x + 757 = 0Δ = 29² - (4 * 757) = 841 - 3028 = - 2187 ← negative value, we stop here
√{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