Answer:
[tex]$ \frac{27}{4x^6y^8} $[/tex]
Step-by-step explanation:
Given:
[tex]$ \frac{4(3x^2y^4)^3}{(2x^3y^5)^4} $[/tex]
[tex]$ (a^b)^c = a^{b.c} $[/tex] and
[tex]$ \frac{x^a}{x^b} = x^{a - b} $[/tex]
We use the above two rules to solve the problem.
So, [tex]$ \frac{4 (3^4 . x^6 . y^{12})}{2^4. x^{12} .y^{20}} $[/tex]
[tex]$ \implies \frac{3^3. x^6. y^{12}}{2^2. x^{12}. y^{20}} $[/tex]
Using the second rule we get:
[tex]$ \frac{3^3}{2^2. x^6. y^{12}}$[/tex]
[tex]$ = \frac{27}{4. x^6. y^8} $[/tex]
Hence, the answer.
