Answer:
Step-by-step explanation:
Prime factorize 9 and 125
9 = 3* 3 = 3²
125 = 5 * 5 * 5 = 5³
3³ * 9² * 125³ ÷ 9³ = [tex]\frac{3^{3}*(3^{2})^{2}*(5^{3})^{3}}{(3^{2})^{3}}\\[/tex]
[tex]= \frac{3^{3}*3^{4}*5^{9}}{3^{6}}\\\\=\frac{3^{3+4}*5^{9}}{3^{6}}\\\\= \frac{3^{7}*5^{9}}{3^{6}}\\\\=3^{7-6}*5^{9}\\=3*5^{9}[/tex]
Exponent laws:
[tex](a^{m})^{n}=a^{m*n}\\\\\\a^{m}*a^{n} = a^{m +n}\\\\\\\frac{a^{m}}{a^{n}}=a^{m-n} \ ; \ m > n[/tex]
