Question 29700
If the exponents(powers) are negative, it puts the term in fraction form;
ex;
{{{x^-2}}}={{{1/x^2}}}
{{{16y^-8}}}={{{1/16^8}}}
So you would have for the equation;
{{{1/x^2}}}+{{{1/16^8}}}
If there were parenthesis involved;
{{{(x)^2}}}+{{{(16y)^8}}}
This means the power is distributed to the whole term;
so in thr term {{{x^2}}}, nothing really changes; but in the term {{{(16y)^8}}}, the 8th power would make the term= {{{16^8}}}*{{{y^8}}}
Hope you understand
=)