Harvard University 🎓 Entrance Exam | Can you solve ?

I did it in a different way:
1000 = 5 x 5 x 5 x 2 x 2 x 2
[x + 1/x ]² = [3^1/2]²
x² + 2 + 1/x² = 3
x² + 1/x² = 1
[x² + 1/x² ]² = 1²
x^4 + 2 + 1/x^4 = 1
x^4 + 1/x^4 = -1
[x^4 + 1/x^4]² = (-1)²
x^8 + 2 + 1/x^8 = 1
x^8 + 1/x^8 = -1
[x^8 + 1/x^8]^5 = (-1)^5
as (a + b)^5 = a^5 + b^5 + 5ab(a³ + b³) + 10a²b²(a + b):
x^40 + 1/x^40 + 5(x^40)(1/x^40)[x^24 + 1/x^24] + 10(x^16)(1/x^16)[x^8 + 1/x^8] = -1
x^40 + 1/x^40 + 5(1)[x^24 + 1/x^24] + 10(1)[-1] = -1 (I)
[x^8 + 1/x^8]³ = (-1)³
x^24 + 1/x^24 + 3(x^8)(1/x^8)[x^8 + 1/x^8] = -1
x^24 + 1/x^24 + 3(1)[-1] = -1
x^24 + 1/x^24 = 2 (II)
(I) and (II):
x^40 + 1/x^40 + 5(1)[x^24 + 1/x^24] + 10(1)[-1] = -1
x^40 + 1/x^40 + 5(1)[2] + 10(1)[-1] = -1
x^40 + 1/x^40 = -1
[x^40 + 1/x^40]^5 = (-1)^5
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[x^200 + 1/x^200]^5 = (-1)^5
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x^1000 + 1/x^1000 = -1

[@ChidexOnlineMathClass01]

Wow, nice. But your method is long