calculate:

(z+1)^6=(z-1)^6

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We have

`(z+1)^6=(z-1)^6`

`(z+1)^6-(z-1)^6=0`

`((z+1)^3)^2-((z-1)^3)^2=0`

`{(z+1)^3+(z-1)^3}{(z+1)^3-(z-1)^3}=0`

`Either`

`(z+1)^3+(z-1)^3=0`

`z^3+3z^2+3z+1+z^3-3z^2+3z-1=0`

`2z^3+6z=0`

`2z(z^2+3)=0`

`2z=0 or z^2+3=0`

`z=0 ,or z^2+3=0`

`z^2=-3`

`z^2=3i^2`

`z=+-isqrt(3)`

`z=0,+-isqrt(3)` (i)

or

`(z+1)^3-(z-1)^3=0`

`z^3+3z^2+3z+1-z^3+3z^2-3z+1=0`

`6z^2+2=0`

`3z^2+1=0`

`3z^2=-1`

`z^2=-(1/3)`

`z^2=i^2/3`

`z=+-isqrt(1/3)`

`z=0,+-isqrt(3),+-isqrt(1/3)`

`Ans.`

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