Find the remaining zero x^3-6x^2+36x-218 zero:-6i

$$x^3-6x^2+36x-216=x^2(x-6)+36(x-6)=(x-6)(x^2+36)=\\\\=(x-6)[x^2-36\cdot i^2)=(x-6)(x-6i)(x+6i)\\\\(x-6)(x-6i)(x+6i)=0\\\\\Leftrightarrow\ \ \ x-6=0\ \ \ \ or\ \ \ \ x-6i=0\ \ \ \ or\ \ \ \ x+6i=0\\\\.\ \ \ \ \ \ x=6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=6i\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-6i$$

I know this isn’t fair, but I happen to know that imaginary or complex roots always
occur in conjugate pairs.   So if -6i is a root, then +6i also must be one.

The expression also has one real root.

It’s between 6.02765006 and 6.02765007.

The reason for that is probably a mis-type or mis-copy in the question.
The ’ 218 ’ at the end was probably supposed to be ’ 216 ’.  In that case,
the real root would have been exactly ’ 6 ’.

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