= Solution
Factor $1-z^3=(1-z)(1+z+z^2)$. The <geometric series> gives, for $|z|<1$,
$$
\frac1{1+z+z^2}=\frac{1-z}{1-z^3}=1-z+z^3-z^4+z^6-z^7+\cdots.
$$
Thus the first three nonzero terms are \b[$1-z+z^3$]. If “first three terms” means the coefficients through degree two, that Taylor polynomial is $1-z+0z^2$. The distinction matters because the quadratic coefficient vanishes. The nearest singularities are the two nontrivial cube <roots of unity>, so the <radius of convergence> is exactly one.
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