Number of perfect squares in $ 4^{6} \cdot 6^{7} \cdot 21^{8}$ Why can't we use this approach:
rewriting, $2^{19} \cdot 7^{8} \cdot 3^{15}$, and then
using $(3\cdot 2)\cdot(4^9)\cdot(49^4)\cdot(9^7)$.
Hence, number of perfect squares is $10\cdot 5\cdot 8$. 
The answer seems to differ. 

I agree with you. If $n=4^{6} \cdot 6^{7} \cdot 21^{8}=2^{19}  \cdot 3^{15}\cdot 7^{8}$ then the number of perfect squares which divide $n$ is
$$(\lfloor 19/2\rfloor+1)\cdot (\lfloor 15/2\rfloor+1)\cdot (\lfloor 8/2\rfloor+1)=10\cdot  8\cdot 5=400.$$
What is the given answer? It is not clear what "Number of perfect squares in" means.
