Ans: A
$\begin{array}{cl}
& i^3(\beta i-3) \\
= & -i(\beta i -3) \\
= & -\beta i^2+3i \\
= & \beta +3i
\end{array}$
$\begin{array}{cl}
& i^3(\beta i-3) \\
= & -i(\beta i -3) \\
= & -\beta i^2+3i \\
= & \beta +3i
\end{array}$
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