How does $2^{12} \times 3^{12} \times 4^6 \times 8^4 \times 9^6 \times 27^4$ become $6^{36}$?
Solution 1:
Break all the numbers down into their prime factors, then recombine.
$2^{12} \times 3^{12} \times 4^6 \times 8^4 \times 9^6 \times 27^4$
= $2^{12} \times 3^{12} \times (2^2)^6 \times (2^3)^4 \times (3^2)^6 \times (3^3)^4$
= $2^{12 + 2 \times 6+ 3 \times 4} \times 3^{12 + 2 \times 6 + 3 \times 4}$
= $2^{36} \times 3^{36}$
= $(2 \times 3)^{36}$
= $6^{36}$