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[@key_board_x]

[(√39 + √3)/√12]² = (√39 + √3)²/(√12)²
[(√39 + √3)/√12]² = [39 + 2√(39 * 3) + 3]/12
[(√39 + √3)/√12]² = [42 + 2√(3 * 13 * 3)]/12
[(√39 + √3)/√12]² = [42 + 6√13]/12
[(√39 + √3)/√12]² = (7 + √13)/2
[(√39 + √3)/√12]⁴ = { [(√39 + √3)/√12]² }²
[(√39 + √3)/√12]⁴ = { (7 + √13)/2 }²
[(√39 + √3)/√12]⁴ = (7 + √13)²/4
[(√39 + √3)/√12]⁴ = (49 + 14√13 + 13)/4
[(√39 + √3)/√12]⁴ = (62 + 14√13)/4
[(√39 + √3)/√12]⁴ = (31 + 7√13)/2
[(√39 + √3)/√12]⁷ = [(√39 + √3)/√12]⁴.[(√39 + √3)/√12]².[(√39 + √3)/√12]
[(√39 + √3)/√12]⁷ = [(31 + 7√13)/2] * [(7 + √13)/2].[(√39 + √3)/√12]
[(√39 + √3)/√12]⁷ = (31 + 7√13).(7 + √13).(√39 + √3)/(2 * 2 * √12)
[(√39 + √3)/√12]⁷ = (217 + 31√13 + 49√13 + 91).(√39 + √3)/(2 * 2 * 2√3)
[(√39 + √3)/√12]⁷ = (308 + 80√13).(√39 + √3)/(2 * 2 * 2√3)
[(√39 + √3)/√12]⁷ = (77 + 20√13).(√39 + √3)/(2√3)
[(√39 + √3)/√12]⁷ = [77√39 + 77√3 + 20√(13 * 39) + 20√(13 * 3)]/(2√3)
[(√39 + √3)/√12]⁷ = [77√(3 * 13) + 77√3 + 20√(13 * 3 * 13) + 20√(13 * 3)]/(2√3) → simpification by √3
[(√39 + √3)/√12]⁷ = [77√13 + 77 + 20√(13 * 13) + 20√13]/2
[(√39 + √3)/√12]⁷ = [97√13 + 77 + 260]/2
[(√39 + √3)/√12]⁷ = [97√13 + 337]/2

[@marcgriselhubert3915]

(Sqrt(39)+sqrt(3))/sqrt(12) = (sqrt(13)+1)/2, so the we are surching ((sqrt(13)+1)^7)/128. That's simpler.
Now, (sqrt(13)+1)^2 = 14+2.sqrt(13); (sqrt(13)+1)^4 = 196+52+56.sqrt(13)= 248+56.sqrt(13)=8.(31+7.sqrt(13)) (a)
sqrt(13+1)^3 = 26+14+16.sqrt(13) = 40+16.sqrt(13) = 8.(5+2.sqrt(13) (b). Then we multiply (a) and (b):
(sqrt(13)+1)^7 = 64.(155+182+97.sqrt(13)) = 64.(337+97.sqrt(13), then finally the result is (337+97.sqrt(13))/2.