Answer
$\dfrac{x^4}{y^4}$
Work Step by Step
RECALL:
(i) $a^m \cdot a^n = a^{m+n}$
(ii) $a^{-m} = \dfrac{1}{a^m}, a\ne0$
(iii) $a^0=1, a\ne0$
Use the rules above to have:
$=x^{5-1}y^{-8-(-4)}z^{3-3}
\\=x^4y^{-8+4}z^{0}
\\=x^4y^{-4}(1)
\\=x^4 \cdot \dfrac{1}{y^4}
\\=\dfrac{x^4}{y^4}$
