How many complex roots does the equation below have? x⁶ + x³ + 1 = 0

Solution:

Given, the equation is x⁶ + x³ + 1 = 0

The equation can be written as (x³)² + x³ + 1 = 0

Let us assume, x³ = t

So, the equation becomes

t² + t + 1 = 0

Now, find the root using the quadratic formula 

t = [-b ± √(b² - 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, x³ = ⍵

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.

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How many complex roots does the equation below have? x⁶ + x³ + 1 = 0

Summary:

The number of complex roots of the equation x⁶ + x³ + 1 = 0 is 6. The number of complex roots is 6 as complex roots always occur in conjugate pairs.