The solutions of x^{6}=1 (x to the power 6 equals 1) are given by x= 1, -1, (-1+√3i)/2, (-1-√3i)/2, (1+√3i)/2, and (1-√3i)/2, where i = √-1 is an imaginary complex number. In this post, we will learn how to find the roots of x^6=1.

## Solutions of x^{6}=1

To solve the given equation x^{6}=1, we need to follow the below steps given here.

**Step 1**:

x^{6}=1

⇒ x^{6 }-1 =0

⇒ (x^{3})^{2} -1^{2} =0

⇒ (x^{3 }-1)(x^{3}+1) =0 [applying the formula of a^{2}-b^{2}=(a-b)(a+b) ]

So either x^{3}-1=0 or x^{3}+1=0

**Step 2**:

Now, we will solve the above two equations.

First, solve x^{3}-1=0

Note we have discussed the solutions of x^{3}-1=0 here. And the solutions are 1, ω, and ω^{2} where ω = (-1+√3i)/2 is the complex number. …(∗)

**Step 3**:

Now, solve x^{3}+1=0

⇒ x^{3 }+1^{3}=0

⇒ (x+1) (x^{2}-x+1) =0 [applying the formula of a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2}) ]

⇒ either x+1=0 or x^{2}-x+1=0

⇒ x=-1

Now, solve x^{2}-x+1=0

Comparing this equation with ax^{2} + bx + c = 0, we get that a=1, b=-1, c=1.

Using Shreedhara Acharya’s formula, the solutions of x^{2}-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 x^{2}-x+1 = 0 are $\dfrac{1 + \sqrt{3}i}{2}$ and $\dfrac{1 – \sqrt{3}i}{2}$.

Thus, the solutions of x^{3}+1=0 are -1, $\dfrac{1 + \sqrt{3}i}{2}$ and $\dfrac{1 – \sqrt{3}i}{2}$. …(∗∗)

**Conclusion**: Combining (∗) and (∗∗), we deduce that the solutions of x^6=1 are ±1, (-1+√3i)/2, (-1-√3i)/2, (1+√3i)/2, and (1-√3i)/2.

So there are only two real roots of x^6=1. Ans there are four complex roots of x^6=1.

## Roots of x^{6}=1

Note that the solutions of the equation x^{6}=1 are the same as the roots of x^{6}=1. Thus, from above we deduce that

the roots of x^{6}=1 are ±1, (-1+√3i)/2, (-1-√3i)/2, (1+√3i)/2, and (1-√3i)/2,

where i = √-1 is the imaginary complex number.

**ALSO READ:**

How to solve quadratic equations

## FAQs

### Q1: What are the solutions of x^{6}=1?

Answer: The solutions of x^6=1 are x=1, -1, (-1+√3i)/2, (-1-√3i)/2, (1+√3i)/2, and (1-√3i)/2, where i = √-1 is an imaginary complex number.

### Q2: What are the roots of x^{6}=1?

Answer: The roots of x^6=1 are 1, -1, (-1+√3i)/2, (-1-√3i)/2, (1+√3i)/2, and (1-√3i)/2.

### Q3: Find the number of real roots of x^{6}=1.

Answer: The number of real roots of x^6=1 is two.

### Q4: Find the number of complex roots of x^{6}=1.

Answer: The number of complex roots of x^6=1 is four.