# How many complex roots does the equation below have? x^{6} + x^{3} + 1 = 0

**Solution:**

Given, the equation is x^{6} + x^{3} + 1 = 0

The equation can be written as (x^{3})^{2} + x^{3} + 1 = 0

Let us assume, x^{3} = t

So, the equation becomes

t^{2} + t + 1 = 0

Now, find the root using the quadratic formula

t = [-b ± √(b^{2} - 4ac)] / 2a

Here, a = 1, b = 1 and c = 1

t = [-(1) ± √((1)^{2} - 4(1)(1))] / 2(1)

t = [-1 ± √(1 - 4)] / 2

t = [-1 ± √-3] / 2

t = (-1 ± 3i) / 2

We know that, (-1 ± 3i) / 2 is a complex number.

⍵ = (-1 ± 3i) / 2

Now, x^{3} = ⍵

Taking cube root,

x = (⍵)^{1/3}

Cube root of a complex number is complex.

Therefore, the number of complex roots is 6 as complex roots always occur in conjugate pairs.

## How many complex roots does the equation below have? x^{6} + x^{3} + 1 = 0

**Summary:**

The number of complex roots of the equation x^{6} + x^{3} + 1 = 0 is 6. The number of complex roots is 6 as complex roots always occur in conjugate pairs.

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