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# Find the complex cube root of 8.

We have to find the complex cube root of 8.

## Answer: The complex cube root of 8 is −1 ± √3i

Let us see how we can find the complex cube root of 8.

**Explanation: **

We can plot the cube roots of 8 in the complex plane on the circle of radius 2.

So, we can write as,

2(cos(0) + isin(0)) = 2

2(cos(2π/3) + isin(2π/3)) = −1 + √3i = 2ω

2(cos(4π/3) + isin(4π/3)) = −1 − √3i = 2ω^{2}

We will find all of the roots of x^{3 }− 8 = 0 to calculate the complex cube root of 8.

x^{3 }− 8 = (x − 2) (x^{2 }+ 2x + 4)

We can solve x^{2 }+ 2x + 4 = 0 using the quadratic formula:

x = −b ± √(b^{2 }− 4ac) / (2a)

= −2 ± √2^{2}−(4 × 1 × 4)/ 2×1

= −2 ± √−12/2

= −1 ± √3i

### Thus, the complex cube root of 8 is given by √8 = −1 ± √3i

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