Question 457202
{{{16bp/(b^2-p^2) - (b-p)/(b+p)}}}


={{{16bp/((b-p)(b+p)) - (b-p)/(b+p)}}}


={{{16bp/((b-p)(b+p)) -((b-p)(b+p))/((b-p)(b+p))}}}


={{{(16bp -(b^2-p^2))/((b-p)(b+p))}}}


={{{(-b^2+16bp-p^2)/((b-p)(b+p))}}}