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558 LearnersLast updated on August 5, 2025


As mentioned above, the cube root of Unity are 1,𝛚, 𝛚², where 1 is a real root, 𝛚 and 𝛚² are the imaginary roots.
The essential features or properties of the cube root of Unity are:
The imaginary roots 𝛚 and 𝛚², when multiplied together, yields 1
𝛚×𝛚²= 𝛚³=1
The summation of the roots is zero → 1+𝛚+𝛚²=0.
The imaginary root 𝛚, when squared, is expressed as 𝛚², which is equal to another imaginary root.
Fact check: Do you know? The values of Cube root of (-1) are -1, -𝛚, and -𝛚²
Now, let us find the meaning of 𝛚 here. To find the cube root of Unity, we will make use of some algebraic formulas. We know that, the cube root of unity is represented as ∛1. Let us assume that ∛1= a so,
∛1= a
⇒ 1 = a3
⇒ a3- 1 = 0
⇒ (a - 1)(a2+a+1) = 0 [using a3-b3= (a - b)(a2+a.b+b2)]
⇒a - 1 =0
⇒ a= 1 …………..(1)
Again, a2+a+1 = 0
⇒ a = (-1 ±√(12–4×1×1)) / 2×1
⇒ a = (-1 ±√(–3)) / 2
⇒ a = (-1 ± i√3) / 2
⇒ a = (-1 + i√3) / 2 …………(2)
Or
a = (-1 - i√3) / 2 …………(3)
From equation (1), (2), and (3), we get,
The roots are → 1, (-1 + i√3) / 2 and (-1 - i√3) / 2
Hence, 𝛚 = (-1 + i√3) / 2
𝛚2= (-1 - i√3) / 2
some common mistakes with their solutions given:


Factorize m²+ mn + n²
We know that, 1+𝛚+𝛚2=0
⇒ 𝛚+𝛚2= -1 ……….(1)
And, 𝛚3=1 …….(2)
So, m2+mn+n2
= m2 - (-1)mn +1× n2
= m2 - (𝛚+𝛚2)mn + 𝛚3× n2 [Using (1) and (2)]
= m2- mn𝛚- mn𝛚2+ n2𝛚3
= m(m-n𝛚) -n𝛚2(m-n𝛚)
= (m-n𝛚)(m-n𝛚2)
Answer : (m-n𝛚)(m-n𝛚2)
We used the properties of the cube root of unity to factorise the expression.
Find 𝛚⁶⁶
𝛚66
=(𝛚3)22
=(1)22
=1
Answer: 1
We used the property 𝛚3=1, and solved the expression.
Prove that (1+𝛚)³+(1+𝛚²)³ = -2
We know that, 1+𝛚+𝛚2=0
⇒1+𝛚= -𝛚2 ……….(1)
And also, 1+𝛚2= -𝛚 ………(2)
LHS = (1+𝛚)3+(1+𝛚2)3
=(-𝛚2)3+(-𝛚)3 [Using (1) and (2)]
=(-𝛚6)+(-𝛚3)
= -(𝛚3)2 - (𝛚3)
=-(1)2 - 1 [using the property 𝛚3=1]
= -1-1
=-2
=RHS [proved]
We proved the given expression to be true using properties of cube root if unity like 1+𝛚+𝛚2=0 and 𝛚3=1.
Prove that (1+𝛚-𝛚²)⁶= -64
We know that, 1+𝛚+𝛚2=0
⇒1+ 𝛚= -𝛚2 ……….(1)
And, 𝛚3=1 …….(2)
LHS
= (1+𝛚-𝛚2)6
=(-𝛚2-𝛚2)6 [using (1)]
=(-2𝛚2)6
=26 × (-𝛚2)6
=64× (-𝛚12)
= 64× (-(𝛚3)4)
= 64× (-(1)4)
= 64× (-1)
= -64
=RHS
LHS=RHS
Hence proved
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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