Table Of Contents
Last updated on November 30th, 2024
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 β1, which actually have three rootsβ 1,π, πΒ², which on multiplication together gives β1β as a product. 1ΓπΓπΒ²=1.
As mentioned above, the cube root of Unity are 1,π, π², where 1 is a real root, π and π² are the imaginary roots.
The essential features or properties of the cube root of Unity are:
The imaginary roots π and π², when multiplied together, yields 1
π×π²= π³=1
The summation of the roots is zero → 1+π+π²=0.
The imaginary root π, when squared, is expressed as π², which is equal to another imaginary root.
Fact check: Do you know? The values of Cube root of (-1) are -1, -π, and -π²
Now, let us find the meaning of π 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 β1. Let us assume that β1= a so,
β1= 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, π = (-1 + i√3) / 2
π2= (-1 - i√3) / 2
Factorize mΒ²+ mn + nΒ²
Find πβΆβΆ
Prove that (1+π)Β³+(1+πΒ²)Β³ = -2
Prove that (1+π-πΒ²)βΆ= -64
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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