Solve x^6-1=0 and Find the Roots

The solutions of x6=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 x6=1

To solve the given equation x6=1, we need to follow the below steps given here.

Step 1:


⇒ x6 -1 =0

⇒ (x3)2 -12 =0

⇒ (x3 -1)(x3+1) =0 [applying the formula of a2-b2=(a-b)(a+b) ]

So either x3-1=0 or x3+1=0

Step 2:

Now, we will solve the above two equations.

First, solve x3-1=0

Note we have discussed the solutions of x3-1=0 here. And the solutions are 1, ω, and ω2 where ω = (-1+√3i)/2 is the complex number. …(∗)

Step 3:

Now, solve x3+1=0

⇒ x3 +13=0

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

⇒ either x+1=0 or x2-x+1=0

⇒ x=-1

Now, 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 $\dfrac{1 – \sqrt{3}i}{2}$.

Thus, the solutions of x3+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 x6=1

Note that the solutions of the equation x6=1 are the same as the roots of x6=1. Thus, from above we deduce that

the roots of x6=1 are ±1, (-1+√3i)/2, (-1-√3i)/2, (1+√3i)/2, and (1-√3i)/2,

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


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Q1: What are the solutions of x6=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 x6=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 x6=1.

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

Q4: Find the number of complex roots of x6=1.

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

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