let a =
let b =

Let S_n = aⁿ+bⁿ     ; S₁ = a+b = 2
Then it satisfies the equation with roots a,b, y²-2y+ab = 0
y² = 2y-ab
S_n+2 = 2S_n+1-ab S_n
S₀ = a⁰+b⁰ = 2
S₂ = 2S₁-abS₀ = 4-2ab
S₃ = 2S₂-abS₁ = 2(4-2ab)-2ab = 8-6ab
While S₃ = 6x+28+28-6x =56
Thereore ab = -8
and
y²-2y-8 = 0
(y-4)(y+2) = 0
y = 4 or -2
When 6x+28 = 4³, 6x = 36, x = 6
When 6x+28 = (-2)³, x = -6
Therefore x = 6
 

another method:

a - b = 2
a³ - b³ = 56
(a-b)(a²+ab+b²) = 56
a²+ab+b² = 28
(a-b)²+3ab = 28
3ab = 24
ab = 8
a+b = = 6
it satisfies the equation
y²6y+8 = 0
(y+4)(y+2) = 0
or (y-4)(y-2) = 0
Same afterwards.

View different answers : Ant