Question 153637
Use the formulas:
{{{1/a^(-m)=a^m}}}
{{{(a^m)^n = a^(mn)}}}
{{{(a^m)(a^n)=a^(m+n)}}}
So
{{{(2^2x^5y^-3z^4)/(4^0x^-2y^-8(z^5)^3)}}}
={{{(2^2x^5y^-3z^4)/(4^0x^-2y^-8z^15)}}}
={{{(2^2/4^0)x^5y^-3z^4x^2y^8z^-15}}}
={{{4x^(5+2)y^(-3+8)z^(4-15)}}}
={{{4x^7y^5z^-11}}}