Answer
$-\dfrac{20x^3}{y^2}$
Work Step by Step
(i) $a^m \cdot a^n = a^{m+n}$
(ii) $\dfrac{a^m}{a^n}=a^{m-n}$
(iii) $a^{-m} = \dfrac{1}{a^m}, a \ne 0, m \gt0$
(iv) $(a^m)^n=a^{mn}$
Use the applicable rules above to find:
$=-20x^{4+(-1)}y^{-3+1}
\\=-20x^3y^{-2}
\\=-20x^3(\frac{1}{y^2})
\\=-\dfrac{20x^3}{y^2}$
