= Solution
{c}
Define the relative reciprocal error $\varepsilon_k=1-px_k$. The iteration has the exact identity
$$
px_{k+1}=3px_k-3(px_k)^2+(px_k)^3=1-(1-px_k)^3,
$$
so
$$
\boxed{\varepsilon_{k+1}=\varepsilon_k^3,\qquad\varepsilon_k=\varepsilon_0^{3^k}.}
$$
Consequently $|1-px_0|<1$ guarantees $x_k\to1/p$. For real $p,x_0$, this condition is $0<px_0<2$. A starting guess can be found without division by choosing $x_0$ with the sign of $p$ and repeatedly halving its magnitude until $0<px_0<2$.
For the absolute signed error $e_k=x_k-1/p$, the same identity gives
$$
\boxed{e_{k+1}=p^2e_k^3.}
$$
Thus the method has <cubic convergence>, of <order of convergence> three, with asymptotic error constant $p^2>0$ for a nonexact convergent start. Convergence requires a suitable guess: $px_0=0$ and $px_0=2$ correspond to $\varepsilon_0=1$ and $\varepsilon_0=-1$, respectively, so both are fixed failures. For $|1-px_0|>1$ the error magnitude grows without bound. The exact initial value $px_0=1$ is already the answer.
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