Answer
The electron configuration of $\mathrm{Au}$ is $[\mathrm{Xe}] 4 f^{14} 5 d^{10} 6 s^{1}$; thus, the electron configuration of $\mathrm{Au}^{3+}$ involves losing the 6s electron and two of the 5d subshell's, yielding the following configuration:
$$[\mathrm{Xe}] 4 f^{14} 5 d^{8}$$
Work Step by Step
See answer above.