    # Cube Roots Of Unity Let the cube root of 1 be x i.e., 3√ 1 = x.

Then by definition, x3 = 1 or x3 � 1 = 0 or (x � 1) (x2 + x + 1) = 0

Either x � 1 = 0 i.e., x = 1 or (x2 + x + 1) = 0

Hence Hence, there are three cube roots of unity which are which the first one is real and the other two are conjugate complex numbers. These complex cube roots of unity are also called imaginary cube roots of unity.

PROPERTIES OF THE CUBE ROOTS OF UNITY:

1. One imaginary cube root of unity is the square of the other. Hence it is clear that one cube root of unity is the square of the other.

Hence if one imaginary cube root of unity be ω, then the other would be ω2.

2. The product of the two imaginary cube roots is 1. Hence the product of the two imaginary cube roots  (Proved)

3. The sum of the three cube roots of unity is zero.

The sum of the three cube roots of unity =  (Proved)

Example 1:

If ω is an imaginary cube root of unity, show that (1 - ω2)(1 � ω4) (1 � ω8)(1 � ω10) = 9

Solution:
L.H.S = (1 - ω2)(1 � ω4)(1 � ω8)(1 � ω10)

= (1 - ω2)(1 � ω)(1 - ω2)(1 � ω) (Since ω4 = ω3. ω = 1. ω = ω; ω8 = ω6. ω2 = 12. ω2 = ω2

and ω10 = ω9. ω = 13. ω = ω )

= (1 - ω2)2(1 � ω )2

= (1 - 2 ω2 + ω4) (1 - 2 ω + ω2)

= (1 + ω - 2 ω2)( 1 + ω2 - 2 ω ) (Since ω4 = ω )

= (1 + ω + ω2 - 3 ω2) ( 1 + ω + ω2 - 3 ω )

= (-3 ω2)(- 3ω ) (Since 1 + ω + ω2 = 0)

= 9 ω3 = 9.1 = 9 (Proved)

Example 2:

Factorize x2 + xy + y2.

Solution:

x2 + xy + y2 = x2 �(-1)xy + 1.y2

= x2 � ( ω + ω2)xy + ω3y2 (Since 1 + ω + ω2 = 0 or ω + ω2 = -1 and ω3= 1)

= x2 � xω � xyω2 + ω3y2

= x(x � yω ) - yω 2(x � yω )

= (x � yω ) (x � yω2)           (Answer).

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