(x + 1/x)^2 = 3
x^2 + 2 + 1/x^2 = 3
x^4 - x^2 + 1 = 0
x = ±i^(1/3), ±i^(5/3)
using i^(1/3), we have
x^37 = i^12 i^(1/3) = i^(1/3)
x^31 = i^10 i^(1/3) = -i^(1/3)
x^47 = i^15 i^(2/3) = -i i^(2/3)
x^26 = i^8 i^(2/3) = i^(2/3)
x^63 = i^21 = i
x^9 = i^3 = -i
Looks like (1-i)i^(2/3) + 6
Kinda funky. Better check my math.
Im trying to find out what the value of this equation is:
x^37+ x^31 + x^47+ x^26+ x^63+ x^9 + 6.
By using the fact that (x + 1/x) ^2 = 3
1 answer
1 answer