Answer
$\pm i, 1$ all with a multiplicity of $1$
Work Step by Step
$P(x)=x^3-x^2+x-1$, factoring the polynomial:
$x^2(x-1)+1(x-1)$,
$(x^2+1)(x-1)$,
$(x^2+1)(x-1)=0$
thus, either $x^2+1=0, x^2=-1, x^2=\pm i$ or $x-1=0, x=1$.
Therefore, the zeros are $\pm i, 1$ all with a multiplicity of $1$
