y+(1/y)=1. find y^3?

Answer 1

y + 1/y = 1
(y + 1/y)³ = 1³
y³ + 3y + 3(1/y)+ 1/y³ = 1
y³ + 3 + 1/y³ = 1
y³ + 2 + 1/y³ = 0
(y³)² + 2y³ + 1 = 0
(y³ + 1)²= 0

Answer: y³ = -1
Note that y itself is imaginary (1 +/- i√3)/2

Answer 2

finding the cube of both sides
y^3 +3y^2*1/y +3y*1/y^2+1/y^3=1
y^3+3(y+1/y)+1/y^3=1
y^3+3+1/y^3=1
call y^3 = z
z+1/z+2=0 z^2+2z+1 =(z+1)^2= 0 so z= y^3=-1
If you go to the given equation
y^2-y+1=0 which has no real roots so you can´t deduce that
y=-1.
y is one of the three complex roots of (-1)^1/3 but not -1

Answer 3

The answer is (-1)^(1/3).

Answer 4

Y+(1/y)=1
1/y =1-y
(1-y)y=1
y-y^2=1
0=y^2-y+1

no real root