Given z^2+2z+4=0, demonstrate z^2 -(8/z)=0?

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You need to test if `z^2 - 8/z` yields `0` , hence, you need to start by bringing the members of expression to a common denominator, such that:

`z^2 - 8/z = 0 => z^3 - 8 = 0 => z^3 - 2^3 = 0`

You need to convert the difference of cubes into the following special product, such that:

`z^3 - 2^3 = (z - 2)(z^2 + 2z + 4)`

Since the problem provides the information that `z^2 + 2z + 4 = 0` , hence, by zero product rule, yields that `(z - 2)(z^2 + 2z + 4) = 0` , thus `z^3 - 8 = 0` .

**Hence, testing if the given identity is `z^2 - 8/z = 0` valid yields that the equivalent expression `z^3 - 8 = 0` is valid, thus, the statement `z^2 - 8/z = 0` holds, under the given conditions.**

Given `z^2+2z+4=0` ,we wish to prove `z^2-8/z=0`

cosider

LHS=`z^2-8/z`

`=(z^3-8)/z`

`=(z^3-2^3)/z`

`=((z-2)(z^2+2z+4))/z`

`=((z-2)xx0)/z` because `z^2+2z+4=0` ( given )

`=0/z`

`=0`

`=RHS`

Hence proved.

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