Remove all unit production rules from
      S \to Aa \mid B &&
      A \to a \mid bc \mid B &&
      B \to A \mid bb
    Note that there are no $\lambda$-productions.\\
    (So no need to first remove $\lambda$-productions.)
    We determine all pairs $A \neq B$ with $A \Rightarrow^+ B$:
      \mpause[1]{S \Rightarrow^+ B} &&
      \mpause{A \Rightarrow^+ B} &&
      \mpause{B \Rightarrow^+ A} &&
      \mpause{S \Rightarrow^+ A}
    Thus we add the following rules:
      S &\to Aa \mid B \alert{\mpause[1]{\mid a}\mpause{\mid bc}\mpause{\mid A}\mpause{\mid bb}} \\
      A &\to a \mid bc \mid B \alert{\mpause{\mid A}\mpause{\mid bb}} \\
      B &\to A \mid bb \alert{\mpause{\mid a}\mpause{\mid bc}\mpause{\mid B}}
    Removing all unit production rules yields the final result:
      S &\to a \mid bb \mid bc \mid Aa &&&
      A &\to a \mid bb \mid bc &&&
      B &\to a \mid bb \mid bc