$(\forall x A(x) \vee \forall x B (x) )\rightarrow \forall x [A(x) \vee B(x)]$
If we found a case which can lead to $T \rightarrow F$ then we can say that it is not valid.
$(\forall x A(x) \vee \forall x B (x) )$ will be true if all x are true for A or all x are true for B or all x are true for Both A and B.
To make $\forall x [A(x) \vee B(x)]$ false we need x which is false for both A and B but this is contradiction because LHS says that all x are true for A or all x are true for B or all x are true for Both A and B.
Hence the given wff is Valid.