Ειναι α^11=α^6×α^3×α^2. α^6=9-4(5)^(1/2) α=[(5)^(1/2)-2]^(1/3) α^3=(5)^(1/2)-2. Αρκει να υπολογισω το α. α^3=(5)^(1/2)-2>0. Αν β^3=(5)^(1/2)+2>0 εχω το συστημα αβ=1 και α^3-β^3=-4........ αβ=1 και α-β=-1. α=[(5)^(1/2)-1]/2 α^2=[3-(5)^(1/2)]/2. Τελικα α^11=[89(5)^(1/2)-199]/2
a>0: a^6=9-4√5 hence a= (√5-1)/2; a^5=(5√5-11)/2 so,
a^11= (89√5-199)/2
E=(89√5-199) /2
9-20/√5 = 9-4√5 = (v5-2)^2 . Again, √5-2 = [1/2(√5-1)]^3. So, 9-20/√5 = [1/2(√5-1)]^6. Thus a = 1/2(√5-1). a^2=1-a gives a^5=5a-3 and a^6=5-8a. So, a^11=89a-55 = 1/2[89√5 -199].
a = ⁶√[9 - (20/√5)] → where: a > 0
a⁶ = 9 - (20/√5)
a⁶ = 9 - [(20√5)/(√5 * √5)]
a⁶ = 9 - [(20√5)/5]
a⁶ = 9 - 4√5
a⁶ = 4 + 5 - 4√5
a⁶ = 4 - 4√5 + 5
a⁶ = (2)² - 2.(2 * √5) + (√5)²
a⁶ = (2 - √5)²
(a³)² = (2 - √5)²
|a³| = |2 - √5|
a³ = √5 - 2 → suppose that:
b³ = √5 + 2
---------------------------------------------------subtraction
a³ - b³ = - 4
a³b³ = (√5 - 2).(√5 + 2)
a³b³ = 5 - 4
a³b³ = 1
ab = 1
Identity:
a³ - b³ = (a - b).(a² + ab + b²) → recall: a³ - b³ = - 4
(a - b).(a² + ab + b²) = - 4
(a - b).[a² + b² + ab] = - 4
(a - b).[(a - b)² + 2ab + ab] = - 4
(a - b).[(a - b)² + 3ab] = - 4 → recall: ab = 1
(a - b).[(a - b)² + 3] = - 4 → let: x = a - b
x.(x² + 3) = - 4
x³ + 3x + 4 = 0 → x = - 1 is an obvious solution
Recall: x = a - b
a - b = - 1 → recall: ab = 1 → b = 1/a
a - (1/a) = - 1
(a² - 1)/a = - 1
a² - 1 = - a
a² + a - 1 = 0
Δ = (1)² - (4 * - 1) = 1 + 4 = 5
a = (- 1 ± √5)/2 → recall the condition: a > 0
a = (- 1 + √5)/2
a = (√5 - 1)/2
a² = [(√5 - 1)/2]²
a² = (√5 - 1)²/2²
a² = (5 - 2√5 + 1)/2²
a² = (6 - 2√5)/2²
a² = (3 - √5)/2
Recall: a³ = √5 - 2 → let's check
a³ = a².a
a³ = [(3 - √5)/2].[(√5 - 1)/2]
a³ = (3 - √5).(√5 - 1)/4
a³ = (3√5 - 3 - 5 + √5)/4
a³ = (4√5 - 8)/4
a³ = √5 - 2 ← it's ok → let's continue
a⁴ = (a²)²
a⁴ = [(3 - √5)/2]²
a⁴ = (3 - √5)²/2²
a⁴ = (9 - 6√5 + 5)/2²
a⁴ = (14 - 6√5)/2²
a⁴ = (7 - 3√5)/2
a⁸ = [(7 - 3√5)/2]²
a⁸ = (7 - 3√5)²/2²
a⁸ = (49 - 42√5 + 45)/2²
a⁸ = (94 - 42√5)/2²
a⁸ = (47 - 21√5)/2
a¹¹ = a⁸.a³
a¹¹ = [(47 - 21√5)/2].(√5 - 2)
a¹¹ = (47 - 21√5).(√5 - 2)/2
a¹¹ = (47√5 - 94 - 105 + 42√5)/2
a¹¹ = (89√5 - 199)/2
a^6 ➖( 9 )^2➖ (20)^2/25=a^6 ➖{ 81 ➖ 400}/25=(a^6)^2 ➖ 319/25={a^36 ➖ 319}/25=283/25=14.3 7^7.3 3^4^3^4.3;1^2^2^1^2^2.3 1^1^1^2.3 2.3(a ➖ 3a+2).
Ειναι α^11=α^6×α^3×α^2. α^6=9-4(5)^(1/2) α=[(5)^(1/2)-2]^(1/3) α^3=(5)^(1/2)-2. Αρκει να υπολογισω το α. α^3=(5)^(1/2)-2>0. Αν β^3=(5)^(1/2)+2>0 εχω το συστημα αβ=1 και α^3-β^3=-4........ αβ=1 και α-β=-1. α=[(5)^(1/2)-1]/2 α^2=[3-(5)^(1/2)]/2. Τελικα α^11=[89(5)^(1/2)-199]/2