Simplify the following; leave all exponents in positive form a. \( \frac{\left(a^{2} b^{1}\right)^{2}}{a^{4} b^{3}} \times\left(\frac{a^{2}}{b^{3}}\right)^{-1} \) b. \( \frac{27^{a}+9^{a}}{3^{3 a}} \) c) \( \frac{8 m^{2} n^{3}}{\left(3 m n^{2}\right)^{3}} \div \frac{\left(4 m^{2}\right)^{2}}{9 m^{5} n^{0}} \) d) \( \left(\frac{x}{y}+\frac{1}{x}\right)^{-2} \) e) \( \frac{30^{p}+96^{p}}{48^{p}} \)

a. To simplify \( \frac{\left(a^{2} b^{1}\right)^{2}}{a^{4} b^{3}} \times\left(\frac{a^{2}}{b^{3}}\right)^{-1} \), we first evaluate the numerator: \((a^{2}b)^{2} = a^{4}b^{2}\). Now, the expression is \(\frac{a^{4} b^{2}}{a^{4} b^{3}} \times b^{3}/a^{2} = \frac{b^{2}}{b^{3}} \times \frac{b^{3}}{a^{2}} = \frac{1}{a^{2}} \Rightarrow a^{-2}\). b. In \( \frac{27^{a}+9^{a}}{3^{3 a}} \), we can rewrite \(27^{a} = (3^3)^{a} = 3^{3a}\) and \(9^{a} = (3^2)^{a} = 3^{2a}\). Thus, the numerator becomes \(3^{3a} + 3^{2a} = 3^{2a}(3^{a}+1)\). This reduces to \(\frac{3^{2a}(3^{a}+1)}{3^{3a}} = \frac{3^{2a}(3^{a}+1)}{3^{2a} \cdot 3^{a}} = \frac{3^{a}+1}{3^{a}} = 1 + \frac{1}{3^{a}}\). c. To simplify \( \frac{8 m^{2} n^{3}}{\left(3 m n^{2}\right)^{3}} \div \frac{\left(4 m^{2}\right)^{2}}{9 m^{5} n^{0}} \), we first find the denominator: \((3mn^2)^3 = 27m^3n^6\). Therefore, we have \(\frac{8m^2n^3}{27m^3n^6}\). Now simplifying gives \(\frac{8}{27} \cdot \frac{1}{m^{1}n^{3}} = \frac{8}{27m n^{3}}\). For the second part, \((4m^2)^2 = 16m^4\), so overall, we simplify this complex fraction to \(\frac{8}{27 m n^{3}} \cdot \frac{9 m^{5}}{16 m^{4}} = \frac{8 \cdot 9}{27 \cdot 16} \cdot \frac{m^{1}}{n^{3}} = \frac{72}{432} \cdot \frac{m^{1}}{n^{3}} = \frac{1}{6} \cdot \frac{m^{1}}{n^{3}} = \frac{m}{6n^{3}}\). d. For \( \left(\frac{x}{y}+\frac{1}{x}\right)^{-2} \), we first combine the terms inside the parentheses with a common denominator: \(\left(\frac{x^2+y}{xy}\right)^{-2} = \left(\frac{xy}{x^2+y}\right)^{2} = \frac{x^{2}y^{2}}{(x^{2}+y)^{2}}\). e. In \( \frac{30^{p}+96^{p}}{48^{p}} \), we convert the terms using their prime factorizations: \(30^{p} = (2 \cdot 3 \cdot 5)^{p}\), \(96^{p} = (2^5 \cdot 3)^{p}\) and \(48^{p} = (2^4 \cdot 3)^{p}\). Thus, \(30^{p} + 96^{p} = 2^{p}(3^p \cdot 5^p + 2^{4p} \cdot 3^p)\). This reduces to: \(\frac{2^{p}(3^{p} (5^{p} + 2^{4p})}{2^{4p} \cdot 3^{p}} = 2^{p - 4p}(5^{p}+2^{4p}) = \frac{1}{2^{3p}}(5^{p}+2^{4p})\).