TWO Easy Math HACKS for Radical Algebra Challenge

[@Fjfurufjdfjd]

χ^3=8(61)^(1/2)-23>0 εστω ψ^3=8(61)^(1/2)+23>0. χ0

[@潘博宇-k4l]

X=[(61)^(1/2)-1]/2. X>0

[@Quest3669]

X= (√61-1)/2>0

[@key_board_x]

(x³ + 23)/8 = √61
x³ + 23 = 8√61
x³ = - 23 + 8√61 → where: x³ > 0 → x > 0
x³ = - 23 + 8√61 → we supose that:
y³ = 23 + 8√61
-------------------------------------------------------------subtraction
x³ - y³ = - 46 ← equation (1)
x³y³ = (- 23 + 8√61).(23 + 8√61)
x³y³ = - 23² + (8√61)²
x³y³ = - 529 + 3904
x³y³ = 3375
x³y³ = 15³
xy = 15
Restart from (1)
x³ - y³ = - 46 → recall: (a³ - b³) = (a - b).(a² + ab + b²)
(x - y).(x² + xy + y²) = - 46
(x - y).[(x² + y²) + xy] = - 46
(x - y).[(x - y)² + 2xy + xy] = - 46
(x - y).[(x - y)² + 3xy] = - 46 → recall : xy = 15
(x - y).[(x - y)² + 45] = - 46 → let : a = (x - y)
a.[a² + 45] = - 46
a³ + 45a + 46 = 0 → a = - 1 ← obvious root → we can factorize: (a + 1)
(a + 1).(a² + za + 46) = 0 → expanding
a³ + za² + 46a + a² + za + 46 = 0 → grouping
a³ + a².(z + 1) + a.(46 + z) + 46 = 0 → compared with: a³ + 45a + 46 = 0
For a² → (z + 1) = 0 → z = - 1
For a → (46 + z) = 45 → z = - 1
Restart
(a + 1).(a² + za + 46) = 0 → where: z = - 1
(a + 1).(a² - a + 46) = 0
First case: (a² - a + 46) = 0
a² - a + 46 = 0
Δ = (- 1)² - (4 * 46) = 1 - 184 = - 183 = 183i²
x³ = (1 ± i√183)/2 ← complex number ← rejected
Second First case: (a + 1) = 0
a = - 1 → recall: a = (x - y)
x - y = - 1 → recall : xy = 15 → y = 15/x
x - (15/x) = - 1
(x² - 15)/x = - 1
x² - 15 = - x
x² + x - 15 = 0
Δ = 1 - (4 * - 15) = 1 + 60 = 61
x = (- 1 ± √61)/2 → recall: x > 0
x = (- 1 + √61)/2 → let's check with: x³ = - 23 + 8√61
x = (- 1 + √61)/2
x³ = x².x
x³ = [(- 1 + √61)/2]².[(- 1 + √61)/2]
x³ = [(- 1 + √61)²/2²].[(- 1 + √61)/2]
x³ = [(1 - 2√61 + 61)/4].[(- 1 + √61)/2]
x³ = [(62 - 2√61)/4].[(- 1 + √61)/2]
x³ = [(31 - √61)/2].[(- 1 + √61)/2]
x³ = (31 - √61).(- 1 + √61)/4
x³ = (- 31 + 31√61 + √61 - 61)/4
x³ = (- 92 + 32√61)/4
x³ = (- 46 + 16√61)/2
x³ = - 23 + 8√61 ← ok