$\displaystyle (18x^3y^4)^{-1}(3x^4y)^4$
The answer I came up with was: $\displaystyle 54x^{13}y$
but I feel this answer is incorrect. Please, what is the correct answer and how does one get it?
Thanks in advance!
$\displaystyle
(18x^3y^4)^{-1}(3x^4y)^4
$
Recall that $\displaystyle (18x^3y^4)^{-1} = \frac {1}{(18x^3y^4)}$
So you would have:
$\displaystyle \frac {81x^{16}y^4}{18x^3y^4}$
Factor out a 9:
$\displaystyle \frac {9x^{16}y^4}{2x^3y^4}$
Factor out an $\displaystyle x^3$:
$\displaystyle \frac {9x^{13}y^4}{2y^4}$
Factor out a $\displaystyle y^4$:
$\displaystyle \frac {9x^{13}}{2}$