# Solve: x^3+i = 0Show complete solution and explain the answer.

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Let write x in the form `re^(it)`

The equation becomes `r^3e^(3it)=-i=e^(-i pi/2)`

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Indentifying moduli and arguments,

`r^3=1` and `3t=-pi/2 +2kpi` for `kinZZ`

`r=1` ( since `r` is a non negative real number) and `t=-\pi/6+2kpi/3`

3 distinct solutions determined by k=0, k=1, k=2` .`

**Solutions :**`x_0=e^(-ipi/6)` , `x_1=e^(ipi/2)=i` , `x_2=e^(i7ipi/6)`

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