<s>[INST] 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. 
 [/INST] As you've noted:
$4^{6}\cdot6^{7}\cdot21^{8}=2^{\color\red{19}}\cdot3^{\color\green{15}}\cdot7^{\color\orange{8}}$

Any perfect square diving this number must be of the following type:
$(2^{0\text{ or }2\text{ or }\dots\text{ or }\lfloor\color\red{19}/2\rfloor\cdot2})\cdot(3^{0\text{ or }2\text{ or }\dots\text{ or }\lfloor\color\green{15}/2\rfloor\cdot2})\cdot(7^{0\text{ or }2\text{ or }\dots\text{ or }\lfloor\color\orange{8}/2\rfloor\cdot2})$

Hence the amount of perfect squares dividing this number is:
$(\lfloor\color\red{19}/2\rfloor+1)\cdot(\lfloor\color\green{15}/2\rfloor+1)\cdot(\lfloor\color\orange{8}/2\rfloor+1)=400$
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