Show that $$ (R ∪ R^{-1})^∗ = R^∗ ∪ R^{−1∗} $$ is false by giving a counterexample.
I tried the following, but every time it keeps coming out as true (instead of false):
If $R = \{(a,b), (a,c)\}$. Then $R^{-1}$ will be $\{(b,a), (c,a)\}$. The union will give: $\{(a,b), (a,c), (b,a), (c,a)\}$. With the ${}^*$, it will be: $\{(a,b), (a,c), (b,a), (c,a), (a,a), (b,b), (c,c)\}$.
Same goes for the right side: $R^*$ will in this case be $\{(a,b), (a,c), (a,a), (b,b), (c,c)\}$. So $R^{-1} = \{(b,a), (c,a)\}$. So the union will be $\{(a,b), (a,c), (b,a), (c,a), (a,a), (b,b), (c,c)\}$.
Which means that it is true (left and right side are equal to each other).
However, when is this false?
Your own example is indeed a counterexample, if you work it out correctly! $(b,c) = (b,a)(a,c)$ is in $(R\cup R^{-1})^*$, but in neither $R^*$ nor $(R^{-1})^*$. (Same with $(c,b)$.)
(In the future, you should remember to define your notation in your questions. It's not clear just from the typographical symbols that $S^*$ is supposed to signify the relation generated by $S$. Indeed, I'm just guessing with the above answer that that's your intent.)