Question 171825
Remember, {{{(x^y)^z=x^(y*z)}}} and {{{(x^n)/(x^m)=x^(m-n)}}}



{{{((x^2y^(-4)z^0)/(3x^6y^(-3)))^4=(x^2y^(-4)z^0)^4/(3x^6y^(-3))^4=(x^(2*4)y^(-4*4)z^(0*4))/(3^4x^(6*4)y^(-3*4))=(x^8y^(-16)z^0)/(81x^24y^(-12))=(1/81)x^(8-24)y^(-16-(-12))=(1/81)x^(-16)y^(-4)=1/(81x^16y^4)}}}




So {{{((x^2y^(-4)z^0)/(3x^6y^(-3)))^4=1/(81x^16y^4)}}}