I'm reading my textbook and it claims that the regular expression $c^*(b \cup (ac)^*)^*$ defines the language $L$ over $\{a,b,c\}$ which consists of all strings that do not contain the substring $bc$.
However, I'm failing to see how that language would contain strings such as $bbbaaa$ or $aaabbb$. Am I missing something or is that regular expression incorrect? The expression I come up with was $ \left( \left( \left( a \cup b \right) ^*a \right) ^* \left( c \cup a \right) ^* \right) ^*b^* $
I checked an older edition of Sudkamp, and it had the same example. Actually the expression is $c^*(b\cup ac^*)^*$, which has a different bracketing. Your $(ac)^*$ specifies strings of the form $acac\dots ac$, whereas Sudkamp has $ac^*$ which means strings like $acc\dots c$.
It follows the arguments of the answer by Raphael, but backwards. Any sequence of $c$'s is either at the beginning of the string, or else it cannot be preceeded by $b$ because there is an explicit $a$ in the expressing.
Here is how you can make sure there is a mistake in the book:
Provided you followed the conversion algorithm correctly and the word is not accepted, there is a mistake in the book: either the claim you quote is wrong or one of the algorithms is wrong.
Spoilers: as mentioned in the comments, the regular expression does not describe the claimed language. The author's epxression enforces "every $a$ is followed by a $c$", which is stronger than the claim.
Your expression looks quite complicated. All you need to do is ensure that every $b$ is followed by $a$, $b$ or the end of the word!
$(b^*a \mid c )^*b^*$
You can also build a suitable automaton -- basically, start with one that accepts everything and modify it to move to a dead state upon reading $bc$ -- and convert it to a regular expression. Perform whatever construction you can more easily prove correct.
