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Given z^2+z+1=0, what is z^8+(1/z^8)?
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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` .
Posted by sciencesolve on September 2, 2013 at 4:23 PM (Answer #1)
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