Question 854955
Maybe the question was two problems:
{{{4x^0y^(-2)z^3/ 4 x}}} and {{{2h^3j^(-3)k^4/3 j^1k^1}}}
 
If negative exponents are acceptable,
{{{4x^0y^(-2)z^3/ 4 x=(4/4)*(x^0/x)*y^(-2)z^3=1*x^(0-1)*y^(-2)z^3=highlight(x^(-1)y^(-2)z^3)}}}
If the teacher wants all exponents to be positive,
I would start by realizing that {{{x^0=1}}} for all {{{x}}} such that {{{x<>0}}} ,
and {{{y^(-2)=1/y^2}}} so
{{{4x^0y^(-2)z^3/ 4 x=(4/4)*(x^0/x)*y^(-2)z^3=1*(1/x)*(1/y^2)*z^3=highlight(z^3/(xy^2))}}}
 
{{{2h^3j^(-3)k^4/3 j^1k^1=(2/3)*h^3*(j^(-3)/j^1)*(k^4/k^1)=(2/3)*h^3*j^(-3-1)*k^(4-1)=(2/3)*h^3*j^(-4)*k^3}}}
OR
{{{2h^3j^(-3)k^4/3 j^1k^1=(2/3)*h^3*(j^(-3)/j^1)*(k^4/k^1)=(2/3)*h^3*j^(-3-1)*k^(4-1)=(2/3)*h^3*j^(-4)*k^3=(2/3)*h^3*(1/j^4)*k^3=2h^3k^3/3 j^4}}}