# Given z^2+z+1=0, what is z^8+(1/z^8)?

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### 1 Answer

You need to perform the following multiplication, such that:

`(z - 1)(z^2 + z + 1) = 0*(z - 1)`

You need to convert the product to the right side into a difference of cubes, such that:

`z^3 - 1 = 0 => z^3 = 1`

You need to evaluate `z^8` , such that:

`z^8 = z^(3+3+2) => z^8 = z^3*z^3*z^2 =>z^8 = 1*1*z^2 => z^8 = z^2`

Replacing `z^2 ` for `z^8` in expression you need to evaluate yields:

`z^8 + 1/z^8 = z^2 + 1/z^2`

The problem provides the information that `z^2 + z + 1 = 0` such that:

`z^2 = -z - 1`

Replacing `- z -1` for `z^2` yields:

`z^2 + 1/z^2 = (-z - 1)^2 + 1/(- z - 1)^2`

`z^2 + 1/z^2 = (z + 1)^2 + 1/(z + 1)^2`

Expanding the squares yields:

`z^2 + 1/z^2 = z^2 + 2z + 1 + 1/(z^2 + 2z + 1)`

Since `z^2 + z + 1 = 0` yields:

`z^2 + 1/z^2 = z + 1/z => z^2 + 1/z^2 = (z^2 + 1)/z`

Since `z^2 + z + 1 = 0 => z^2 + 1 = -z` , such that:

`z^2 + 1/z^2 = (-z)/z `

Reducing duplicate factors yields:

`z^2 + 1/z^2 = -1`

**Hence, evaluating the expression `z^8 + 1/z^8` , under the given conditions, yields `z^8 + 1/z^8 = -1` .**

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