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484 LearnersLast updated on August 5, 2025

The cube root of Unity (or one) are the values which, when multiplied together, gives the original number Unity. The Cube Root of Unity is represented by โ1, which actually have three rootsโ 1,๐, ๐ยฒ, which on multiplication together gives โ1โ as a product. 1ร๐ร๐ยฒ=1.
As mentioned above, the cube root of Unity are 1,๐, ๐², where 1 is a real root, ๐ and ๐² are the imaginary roots.
The essential features or properties of the cube root of Unity are:
The imaginary roots ๐ and ๐², when multiplied together, yields 1
๐×๐²= ๐³=1
The summation of the roots is zero → 1+๐+๐²=0.
The imaginary root ๐, when squared, is expressed as ๐², which is equal to another imaginary root.
Fact check: Do you know? The values of Cube root of (-1) are -1, -๐, and -๐²
Now, let us find the meaning of ๐ 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 โ1. Let us assume that โ1= a so,
โ1= 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, ๐ = (-1 + i√3) / 2
๐2= (-1 - i√3) / 2
some common mistakes with their solutions given:
Factorize mยฒ+ mn + nยฒ
We know that, 1+๐+๐2=0
⇒ ๐+๐2= -1 ……….(1)
And, ๐3=1 …….(2)
So, m2+mn+n2
= m2 - (-1)mn +1× n2
= m2 - (๐+๐2)mn + ๐3× n2 [Using (1) and (2)]
= m2- mn๐- mn๐2+ n2๐3
= m(m-n๐) -n๐2(m-n๐)
= (m-n๐)(m-n๐2)
Answer : (m-n๐)(m-n๐2)
We used the properties of the cube root of unity to factorise the expression.
Find ๐โถโถ
๐66
=(๐3)22
=(1)22
=1
Answer: 1
We used the property ๐3=1, and solved the expression.
Prove that (1+๐)ยณ+(1+๐ยฒ)ยณ = -2
We know that, 1+๐+๐2=0
⇒1+๐= -๐2 ……….(1)
And also, 1+๐2= -๐ ………(2)
LHS = (1+๐)3+(1+๐2)3
=(-๐2)3+(-๐)3 [Using (1) and (2)]
=(-๐6)+(-๐3)
= -(๐3)2 - (๐3)
=-(1)2 - 1 [using the property ๐3=1]
= -1-1
=-2
=RHS [proved]
We proved the given expression to be true using properties of cube root if unity like 1+๐+๐2=0 and ๐3=1.
Prove that (1+๐-๐ยฒ)โถ= -64
We know that, 1+๐+๐2=0
⇒1+ ๐= -๐2 ……….(1)
And, ๐3=1 …….(2)
LHS
= (1+๐-๐2)6
=(-๐2-๐2)6 [using (1)]
=(-2๐2)6
=26 × (-๐2)6
=64× (-๐12)
= 64× (-(๐3)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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