= Solution
If $x$ is invertible, part (i)'s inverse calculation gives an inverse for $x^*$, and therefore both $xx^*$ and $x^*x$ are invertible. Conversely, if both products are invertible, set
$$
r=x^*(xx^*)^{-1},\qquad \ell=(x^*x)^{-1}x^*.
$$
Then $xr=1$ and $\ell x=1$: $r$ is a <right inverse of an algebra element> $x$ and $\ell$ is a <left inverse of an algebra element> $x$. They coincide because $\ell=\ell(xr)=(\ell x)r=r$, providing a two-sided inverse. Thus
$$
\boxed{x\text{ invertible}\iff xx^*\text{ and }x^*x\text{ both invertible},\qquad x^{-1}=x^*(xx^*)^{-1}=(x^*x)^{-1}x^*.}
$$
Both products are required in a general <Banach algebra>; a single one-sided inverse need not suffice.
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