Find the possible real values of 1/√3 ( i^1/3 + i^-1/3)

Question

A−2,0, and 2

B−1,0, and 1

C−2√3,0, and 2√3

D−1√3,0, and 1√3



E−13,0, and 13

Answer 1

(1/sqr3)(i^(1/3+i^-(1/3)= (1/sqr3)(i+1/i)=(1/sqr3)(-1+1)/i= 0

cube root of i =i

but there are 3 cube roots

x=cube root of i

x^3 =i

x^6=-1

x^6+1=0

(x^2+1)(x^4-x^2+1)=0

x^2=.5+.5isqr3, .5-.5isqr3

x=sqr(.5+.5sqr3), sqr(.5-.5isqr3)

(1/sqr3)sqr(.5+/-.5isqr3)

=sqr(.5/sqr3 +/-.5i)

C or D look most likely given the sqr3 in them