Answer:
2) [tex]\displaystyle -9a^3b^5[/tex]
1) [tex]\displaystyle \frac{x^8}{3}[/tex]
Step-by-step explanation:
2) [tex]\displaystyle \frac{(27a^3b^3)(-2ab^4)}{6ab^2} \Rightarrow \frac{(3ab)^3(-2ab^4)}{6ab^2} \\ \\ \boxed{-9a^3b^5} \rightarrow \frac{-54a^4b^7}{6ab^2}[/tex]
1) [tex]\displaystyle \frac{16x^{10}y^2}{48x^2y^2} \Rightarrow \frac{(-4x^5y)^2}{48x^2y^2} \\ \\ \boxed{\frac{x^8}{3}}[/tex]
The Quotient-to-Power Rule states that whenever you are dividng like-terms, you keep the coefficients and deduct the exponents.
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