Nice solution but requires a lot of calculation. I solved it another way
(a-2)/b + (b-2)/a = 1
Find a, b where a,b are integers.
Observation a≠ 0, b≠0
Multiply both sides of equation by ab
We get
a²-2a+b²-2b= ab→
a²-ab-2a+b²-2b=0
Multiply both sides by 2
2a²-2ab-4a+2b²-4b=0→
(a²-2ab+b²)+a²-4a+b²-4b=0→
(a-b)²+(a²-4a+4)+(b²-4b+4)-8=0→
(a-b)²+(a-2)²+(b-2)²=8
Eq. 1
Observation
1. a and b are positive intgers otherwise LHS will be greater than 9 (a-2 or b-2 willbe less than -3)
2. Any of (a-b), (a-2) or (b-2) is not greater than 2
So a-2≤2 →a≤4
 Similarly b-2≤2→b≤4
Case 1
a=1
Then from Eq 1
(1-b)²+(1-2)²+(b-2)²=8→
1-2b+b² +1+b²-4b+4=8→
2b²-6b-2=0→
b²-3b-1=0
D=(-3)²-4(1)(-1)=13 not a perfect square.
No integer solution
Rejected
Case 2
a=2
(2-b)²+(2-2)²+(b-2)²=8→
(2-b)²+(b-2)²=8→
{(-1)(b-2)}²+(b-2)²=8→
2(b-2)²=8→
(b-2)²=4→
b-2=±2→
b= 0 or b=4
b= 0 is rejected
So we have
(a,b)= (2,4)
Case 3
a=3
(3-b)²+(3-2)²+(b-2)²=8→
9-6b+b² +1+b²-4b+4=8→
2b²-10b+6=0→
b²-5b+3=0
D= (-5)²-4(1)(3)=13
Not a perfect square
Rejected
Case 4
(4-b)²+(4-2)²+(b-2)²=8→
16-8b+b²+4+b²-4b+4=8→
2b²-12b+16=0→
b²-6b+8=0→
(b-4)(b-2)=0→
b=4 or b=2
So (a,b)=(4,4),(4,2)
Finally we have
(a,b)=(2,4),(4,4),(4,2)

x⁹+1/x⁹=(x+1/x){x⁸+1/x⁸-(x⁶+1/x⁶)+(x⁴+1/x⁴)-(x²+1/x²)＋1}=5778
2⁹=512　3⁹=19638　　2＜x＜3　x+1/x=t　2＜t＜4
x²+1/x=t²-2　x³+1/x³=t³-3t　x⁴+1/x⁴=t⁴-4t²+2
x⁶+1/x⁶=(t³-3t)²-2=t⁶-6t⁴+9t²-2
x⁸+1/x⁸=(x⁴+1/x⁴)-2=t⁸-8t⁶+20t⁴-16t²+2
x⁹+1/x⁹=t(t⁸-9t⁶+27t⁴-30t²+9)=5778=3×1926=t×f(t)
Assume t is 3　
f(3)=3⁸-9×3⁶+27×3⁴-30×9+9=19683-19683+2187-270+9=1926
t=3　x+1/x=3　x²-3x+1=0　x=(3±√5)/2

[x³ + (1/x³)]³ = [x³ + (1/x³)]².[x³ + (1/x³)]
[x³ + (1/x³)]³ = [x⁶ + 2 + (1/x⁶)].[x³ + (1/x³)]
[x³ + (1/x³)]³ = x⁹ + x³ + 2x³ + (2/x³) + (1/x³) + (1/x⁹)
[x³ + (1/x³)]³ = x⁹ + (1/x⁹) + 3x³ + (3/x³)
[x³ + (1/x³)]³ = x⁹ + (1/x⁹) + 3.[x³ + (1/x³)] → given: x⁹ + (1/x⁹) = 5778
[x³ + (1/x³)]³ = 5778 + 3.[x³ + (1/x³)]
[x³ + (1/x³)]³ - 3.[x³ + (1/x³)] = 5778 → let: a = x³ + (1/x³)
a³ - 3a = 5778 → let's find a number a such as: a³ > 5778 → a = 18 → a³ = 18³ = 5832 → let's check it
5832 - 3a = 5778
3a = 5832 - 5778
3a = 54
a = 18 ← ok → recall: a = x³ + (1/x³)
x³ + (1/x³) = 18 → recall: x³ + (1/x³) = [x + (1/x)]³ - 3.[x + (1/x)]
[x + (1/x)]³ - 3.[x + (1/x)] = 18 → let: b = x + (1/x)
b³ - 3b = 18 → → let's find a number b such as: b³ > 18 → b = 3 → b³ = 27 → let's check it
27 - 3b = 18
3b = 27 - 18
3b = 9
b = 3 ← ok → recall: b = x + (1/x)
x + (1/x) = 3
(x² + 1)/x = 3
x² + 1 = 3x
x² - 3x + 1 = 0
Δ = (- 3)² - 4 = 5
x = (3 ± √5)/2

Note that 5778 is the 18th Lucas number. x+1/x = 3 = the 2nd Lucas number. Finally x = the square of either phi (the golden ratio) or its conjugate.

Thank you for the help truly valuable video

Can you explain your approach as to how you rewrite the equations after you find a solution?

x⁹ + 1/x⁹ = 5778
x⁹ + x⁻⁹ = 5778 = ϕ¹⁸ + ϕ⁻¹⁸
x⁹ = {ϕ¹⁸ , ϕ⁻¹⁸}
x = {ϕ² * exp{2*π*j*k/9), ϕ⁻² * exp{2*π*j*k/9)} || k ∈ Z
x ∩ R = {ϕ², ϕ⁻²} = (3 ± √5)/2

we notice that the 18 solutions lie on 2 concentric enneagons w/ radius ϕ² & ϕ⁻²...looks pretty...but im too lazy to multiply all of them out...

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3

(x+1/x)^3=x^3+1/x^3+3(x+1/x)
x+1/x =∂
x^3+1/x^3 =∂^3-3∂
(x^3+1/x^3)^3=x^9+1/x^9+3(x^3+
    1/x^3)
 x^9+1/x^9=(x^3+1/x^3)^3-3(x^3+
   1/x^3)
x^3+1/x^3=β
x^9+1/x^9 =β^3-3β =3^3×2×107
     β^3=3(β+3^2×2×107)
    β =3t
27t^3=9(t+3×2×107)
3t^3=t.+3×2×107
     t=3m
(3m)^3-m=2×107
m(27m^2-1)=2×107
    m=2
LHS=2×(27×4-1)=2×107
∴LHS=RHS      m=2
 t=3×2=6
β=3×6=18
∂^3-3∂-18=0
(∂-3)｛(∂+3/2)^2+15/4｝=0
     ∴∂=3
x+1/x=3
x^2-3x+1=0
 x=(3±√5)/2

x^9 +1/x^9=5778
x=?
(x^3 +1/x^3)^3 -3(x^3 +1/x^3)=5778
Let x^3 +1/x^3=y
y^3 -3y -5778=0
y^3 -18y^2 +18y^2 -324y +321y -5778=0
321*18=2568+3210=5778
(y^2)(y-18) +18y(y-18) +321(y-18)=0
y1=18
Delta<0
x^3 +1/x^3=18
(x +1/x)^3 -3(x+1/x)=18
Let x+1/x=z
z^3 -3z -18=0
z^3 -3z^2 +3z^2 -9z +6z -18=0
(z^2)(z-3) +3z(z-3) +6(z-3)=0
z1=3
Delta=9-24<0
x +1/x=3
x^2 +1=3x
x^2 -3x +1=0
x=(3+-sqrt(9-4))/2
x=(3+-sqrt(5))/2
x1=(3+sqrt(5))/2
x2=(3-sqrt(5))/2