Answer:
➤ Solution :-
{\tt \leadsto \dfrac{{12}^{4} \times {9}^{3} \times 4}{{6}^{3} \times {8}^{2} \times 27}}⇝
6
3
×8
2
×27
12
4
×9
3
×4
{\tt \leadsto \dfrac{{12}^{4} \times {9}^{3} \times {2}^{2}}{{6}^{3} \times {8}^{2} \times {3}^{3}}}⇝
6
3
×8
2
×3
3
12
4
×9
3
×2
2
{\tt \leadsto \dfrac{({3}^{1} \times {2}^{2}{)}^{4} \times ({3}^{2}{)}^{3} \times {2}^{2}}{{(2 \times 3)}^{3} \times {({2}^{3})}^{2} \times {3}^{3}}}⇝
(2×3)
3
×(2
3
)
2
×3
3
(3
1
×2
2
)
4
×(3
2
)
3
×2
2
{\tt \leadsto \dfrac{{3}^{1 \times 4} \times {2}^{2 \times 4} \times {3}^{2 \times 3} \times {2}^{2}}{{2}^{3} \times {{3}^{3}} \times {2}^{3 \times 2} \times {3}^{3}}}⇝
2
3
×3
3
×2
3×2
×3
3
3
1×4
×2
2×4
×3
2×3
×2
2
{\tt \leadsto \dfrac{{3}^{4} \times {2}^{8} \times {3}^{6} \times {2}^{2}}{{2}^{3} \times {{3}^{3}} \times {2}^{6} \times {3}^{3}}}⇝
2
3
×3
3
×2
6
×3
3
3
4
×2
8
×3
6
×2
2
{\tt \leadsto \dfrac{{3}^{4 + 6} \times {2}^{8 + 2}}{{2}^{3 + 6} \times {{3}^{3 + 3}}}}⇝
2
3+6
×3
3+3
3
4+6
×2
8+2
{\tt \leadsto \dfrac{{3}^{10} \times {2}^{10}}{{2}^{9} \times {{3}^{6}}}}⇝
2
9
×3
6
3
10
×2
10
{\tt \leadsto {3}^{10 - 6} \times {2}^{10 - 9}}⇝3
10−6
×2
10−9
{\tt \leadsto {3}^{4} \times {2}^{1}}⇝3
4
×2
1
{\tt \leadsto 3 \times 3 \times 3 \times 3 \times 2}⇝3×3×3×3×2
{\tt \leadsto 81 \times 2 = \boxed{\tt 162}}⇝81×2=
162
\Huge\therefore∴ The answer is 162.
━━━━━━━━━━━━━━━━━━━━━━
\dashrightarrow⇢ Some related equations :-
{\sf \longrightarrow {a}^{m} \times {a}^{n} = {a}^{m + n}}⟶a
m
×a
n
=a
m+n
{\sf \longrightarrow {a}^{m} \div {a}^{n} = {a}^{m - n}}⟶a
m
÷a
n
=a
m−n
{\sf \longrightarrow {({a}^{m})}^{n} = {a}^{m \times n}}⟶(a
m
)
n
=a
m×n
{\sf \longrightarrow {a}^{ - n} = \dfrac{1}{ {a}^{n}}}⟶a
−n
=
a
n
1
{\sf \longrightarrow {a}^{0} = 1}⟶a
0
=1