Unit-I: Sets and Functions
Unit-II: Algebra
Unit-III: Coordinate Geometry
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability

Properties of the cube roots of unity

Let  x^3-1=(x-1)(x^2+x+1)=0 where (x-1)=0,~ (x^2+x+1)=0

\implies x=1, ~ \alpha=\dfrac{-1+\sqrt{3}i}{2}, ~~ \beta=\dfrac{-1-\sqrt{3}i}{2}

Which have some important facts:

(i)  \overline{\alpha}=\beta,~~ \overline{\beta}=\alpha, i.e., the two complex roots of the equation are conjugate of each other.

(ii)  \alpha^2=\beta,~~ \beta^2=\alpha, i.e., the square root of any one complex roots is the other complex root.

(iii) If we denote  w to be one complex root, then the three roots are 1, w, w^2 .

(iv) Since  w  is a root of equation  x^3=1, we obtain  w^3=1, where  w^4=w , w^5=w^2,w^6=(w^3)^2=1 i.e., w^{3m}=1, w^{3m+1}=w, w^{3m+2}=w^2 , m \in N .

(v) 1+w+w^2=0 , i.e, the sum of the three cube roots of unity vanishes.

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