Question 381979
{{{x^8y^(-14)/x^(-2)y^12}}}
The rule for exponents when dividing is to subtract the exponents. So
{{{x^8y^(-14)/x^(-2)y^12 = x^(8-(-2))y^(-14-12) = x^10y^(-26)}}}
If you don't want negative exponents in your answer, then:
{{{x^10y^(-26) = x^10/y^26}}} 
since {{{a^(-n) = 1/a^n}}}