5. Simplify the following, write the final answer with positive exponents a) \( \quad\left(a^{2} b\right)^{3} \times \frac{a^{4} b}{\left(a b^{3}\right)^{-1}} \) b) \( \quad\left(a b^{3}\right)^{0} \) c) \( \quad \frac{x^{5} y^{-3}}{\left(x^{2} y^{2}\right)^{2}} \times\left(\frac{1}{x y^{2}}\right)^{-1} \) d) \( \quad \frac{8 m^{2} n}{9 m^{-3} n^{2}} \div \frac{16 m^{3} n^{-1}}{3 m^{4} n} \) e) \( \quad \frac{a b^{2} c}{\left(a b c^{2}\right)^{2}} \times \frac{\left(a^{-2} b^{3} c\right)^{-1}}{a^{0} b^{4}} \) f) \( \quad 2^{3} a^{3} b^{4} \times 3^{2} a^{-2} b^{2} \) g) \( \quad \frac{p^{3} q^{2}}{p^{-2} q^{3}} \times \frac{16 p^{3} q^{3}}{\left(2 p^{-1} q^{2}\right)^{3}} \) h) \( \left(\frac{m^{3} n}{m n^{2}}\right)^{-1}+\frac{m^{3} n^{2}}{n} \) i) \( \quad\left(\frac{x}{y}+\frac{y^{2}}{x^{2}}\right)^{-1} \)

**a)** We start with \[ \left(a^{2}b\right)^3 = a^{2\cdot 3}b^3 = a^6 b^3. \] Also, note that \[ \frac{a^4b}{(ab^3)^{-1}} = a^4b \times (ab^3) = a^{4+1}b^{1+3} = a^5b^4. \] Multiplying these results gives \[ a^6b^3 \times a^5b^4 = a^{6+5}b^{3+4} = a^{11}b^7. \] --- **b)** Any nonzero number raised to the 0th power equals 1: \[ \left(ab^3\right)^0 = 1. \] --- **c)** Rewrite the expression: \[ \frac{x^5 y^{-3}}{\left(x^2 y^2\right)^2} \times \left(\frac{1}{xy^2}\right)^{-1}. \] First, simplify the denominator: \[ \left(x^2y^2\right)^2 = x^{2\cdot2}y^{2\cdot2} = x^4y^4. \] Thus, \[ \frac{x^5 y^{-3}}{x^4y^4} = x^{5-4}y^{-3-4} = xy^{-7}. \] Next, \[ \left(\frac{1}{xy^2}\right)^{-1} = xy^2. \] Multiplying: \[ xy^{-7} \times xy^2 = x^2y^{-7+2} = x^2y^{-5}. \] Expressing with positive exponents: \[ \frac{x^2}{y^5}. \] --- **d)** The expression is \[ \frac{8m^2n}{9m^{-3}n^2} \div \frac{16m^3n^{-1}}{3m^4n}. \] First, simplify the left fraction: \[ \frac{8m^2n}{9m^{-3}n^2} = \frac{8}{9}m^{2-(-3)}n^{1-2} = \frac{8}{9}m^5n^{-1} = \frac{8m^5}{9n}. \] Next, simplify the right fraction: \[ \frac{16m^3n^{-1}}{3m^4n} = \frac{16}{3}m^{3-4}n^{-1-1} = \frac{16}{3}m^{-1}n^{-2} = \frac{16}{3m\,n^2}. \] Now divide: \[ \frac{8m^5}{9n} \div \frac{16}{3m\,n^2} = \frac{8m^5}{9n} \times \frac{3m\,n^2}{16} = \frac{8\cdot3}{9\cdot16} m^{5+1}n^{-1+2}. \] This simplifies as: \[ \frac{24}{144} m^6 n = \frac{1}{6}m^6n. \] --- **e)** We have \[ \frac{ab^2c}{(abc^2)^2} \times \frac{(a^{-2}b^3c)^{-1}}{a^0b^4}. \] First, compute \[ (abc^2)^2 = a^2b^2c^4, \] so \[ \frac{ab^2c}{a^2b^2c^4} = a^{1-2}b^{2-2}c^{1-4} = a^{-1}c^{-3}. \] Next, \[ (a^{-2}b^3c)^{-1} = a^2b^{-3}c^{-1}. \] Since \( a^0 = 1 \), the second part becomes: \[ \frac{a^2b^{-3}c^{-1}}{b^4} = a^2b^{-3-4}c^{-1} = a^2b^{-7}c^{-1}. \] Multiplying the two results: \[ a^{-1}c^{-3} \times a^2b^{-7}c^{-1} = a^{-1+2}b^{-7}c^{-3-1} = a^1b^{-7}c^{-4}. \] With positive exponents, this is: \[ \frac{a}{b^7c^4}. \] --- **f)** Multiply: \[ 2^3a^3b^4 \times 3^2a^{-2}b^2 = 8a^3b^4 \times 9a^{-2}b^2 = 72a^{3-2}b^{4+2} = 72ab^6. \] --- **g)** We begin with \[ \frac{p^3q^2}{p^{-2}q^3} \times \frac{16p^3q^3}{\left(2p^{-1}q^2\right)^3}. \] First fraction: \[ \frac{p^3q^2}{p^{-2}q^3} = p^{3-(-2)}q^{2-3} = p^5q^{-1}. \] Now, expand the denominator in the second fraction: \[ \left(2p^{-1}q^2\right)^3 = 2^3 (p^{-1})^3 (q^2)^3 = 8p^{-3}q^6. \] Thus, the second fraction becomes: \[ \frac{16p^3q^3}{8p^{-3}q^6} = 2p^{3-(-3)}q^{3-6} = 2p^6q^{-3}. \] Multiplying the two results: \[ p^5q^{-1} \times 2p^6q^{-3} = 2p^{5+6}q^{-1-3} = 2p^{11}q^{-4}. \] Expressed with positive exponents: \[ \frac{2p^{11}}{q^4}. \] --- **h)** The expression is \[ \left(\frac{m^3n}{mn^2}\right)^{-1} + \frac{m^3n^2}{n}. \] Simplify the fraction inside the parentheses: \[ \frac{m^3n}{mn^2} = m^{3-1}n^{1-2} = m^2n^{-1}. \] Taking the \(-1\) power: \[ \left(m^2n^{-1}\right)^{-1} = m^{-2}n. \] Now, simplify the second term: \[ \frac{m^3n^2}{n} = m^3n^{2-1} = m^3n. \] Thus, the sum is: \[ m^{-2}n + m^3n = n\left(m^{-2} + m^3\right).