Question

(3y^0 x^2)^6 times y^-3 zx8

Answer 1

Ok so some basic rules

$(x^m)(x^n)=x^{m+n}$
$(ab)^c=(a^c)(b^c)$
$(x^m)^n=x^{mn}$
$x^0=1$
$x^{-m}= \frac{1}{x^m}$

so
$[(3y^0x^2)^6][y^{-3}zx^8]$
$[(3(1)x^2)^6][y^{-3}zx^8]$
$[(3x^2)^6][y^{-3}zx^8]$
$[(3^6)((x^2)^6)][y^{-3}zx^8]$
$[(3^6)(x^{12})][y^{-3}zx^8]$
$(3^6)(x^{12})(y^{-3})(z)(x^8)$
$(3^6)(x^{12})(x^8)(y^{-3})(z)$
$(3^6)(x^{20})(\frac{1}{y^3})(z)$
$\frac{3^6x^{20}z}{y^3}$
$\frac{729x^{20}z}{y^3}$