this question asked so many time in final year exam ,, but we were not able to solve this...... If 1,w,w^2 are the cube roots of unity, then find the roots of (z – 2)3 + 27 = 0.
If:
$\displaystyle (z-2)^3=-27=27 e^{(\pi+2n\pi)i},\ n=0, \pm1,\ \pm 2,\ ..$
so:
$\displaystyle (z-2)=3 e^{(\pi+2n\pi)i/3},\ n=0, \pm1,\ \pm 2,\ ..$
but only three of the exponential terms on the right are distinct and these are the cube roots of $\displaystyle -1$, and the three distinct values are $\displaystyle -1$, $\displaystyle -w$, $\displaystyle -w^2$, from which the previous result follows.
This uses Eulers formula for complex numbers and the laws of powers.
CB