Question 256447
We are given:
(i) {{{(-7a^2b^3) (-14ab^-3)}}}
First, multiply -7 and -14 to get 
(ii) {{(98a^2b^3) (ab^-3)}}}
Second, multiply the a's to get
(iii) {{(98a^3b^3) (b^-3)}}}
Third, multiply the b's to get
(iv) {{{98a^3b^0}}}
However, b^0 = 1, so we get
{{{98a^3}}}