# x^3=1 Solution | x^3=1 Roots

The solutions of x3=1 (x cubed equals 1) are given by x= 1, ω, and ω2 where ω = (-1+√3i)/2 is the complex number. In this post, we will learn how to find the root of x3=1.

## Solutions of x3=1

The given equation is x3=1.

Step 1: Apply the formula of a3-b3.

⇒ x3 -1 =0

⇒ x3 -13 =0

⇒ (x -1)(x2+x+1) =0 [applying the formula of a3-b3=(a-b)(a2+ab+b2) ]

So either x-1=0 or x2+x+1=0 …(∗)

Step 2: Now, we will solve the above two equations.

x-1=0 ⇒ x=1.

and solve x2+x+1=0

Comparing this equation with ax2 + bx + c = 0, we get that a=1, b=1, c=1.

Using Shreedhara Acharya’s formula, the solutions of x2+x+1=0 are given by

x = $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

= $\dfrac{-1 \pm \sqrt{-3}}{2}$

= $\dfrac{-1 \pm \sqrt{3}i}{2}$.

Therefore, the solutions of x2+x+1 = 0 are $\dfrac{-1 + \sqrt{3}i}{2}$ and $\dfrac{-1 – \sqrt{3}i}{2}$.

Step 3: Set ω = (-1+√3i)/2.

Then ω2 = (-1-√3i)/2.

So the solutions of x2+x+1 = 0 are ω and ω2.

Therefore, from (∗), we conclude that the solutions of x3=1 are 1, ω, and ω2 where ω = (-1+√3i)/2.

## Roots of x3=1

As the solutions of the equation x3=1 are the roots of x3=1, from above we deduce that the roots of x3=1 are

1, ω, and ω2

where ω = (-1+√3i)/2 is a complex number.

How to solve linear equations

Solve x2+25=0

## FAQs

### Q1: What are the solutions of x^3=1?

Answer: 1, ω, and ω2 are the solutions of x3=1, where ω = (-1+√3i)/2 is the complex number.

### Q2: What are the roots of x^3=1?

Answer: 1, ω, and ω2 are the roots of the cubic equation x3=1, where ω = (-1+√3i)/2.