An algebra problem by Anshuman Shreshth

$y+\frac{1}{y}=1$

or, $y^2-y+1=0$

or, $(y+1)(y^2-y+1)=0$

or, $y^3+1=0$

or, $y^3=\boxed{-1}$

y+1/y=1 i.e y^2-y+1=0 i.e y=(1+i(3^(1/2)))/2 or y=(1-i(3^(1/2)))/2
Take any value of y (both produce same answer) and cube it