1. **Stating the problem:** Simplify the given expressions involving roots and rational exponents. 2. **Important rules:** - $\sqrt[n]{a^m} = a^{\frac{m}{n}}$ converts roots to rational exponents. - When multiplying like bases, add exponents: $a^x \cdot a^y = a^{x+y}$. - When dividing like bases, subtract exponents: $\frac{a^x}{a^y} = a^{x-y}$. - Powers of powers multiply exponents: $(a^x)^y = a^{xy}$. --- **Part a)** Simplify $\sqrt{2a^3} \cdot \sqrt{18a}$ $$\sqrt{2a^3} \cdot \sqrt{18a} = \sqrt{2a^3 \cdot 18a} = \sqrt{36a^{4}} = 6a^{2}$$ Explanation: Multiply under one root, simplify $36$ and $a^{4}$. --- **Part b)** Simplify $3 \cdot \sqrt[4]{x^3} \cdot 4 \cdot \sqrt[4]{x^9} \cdot 2 \cdot \sqrt[4]{x^5}$ Rewrite roots as exponents: $$3 \cdot 4 \cdot 2 \cdot x^{\frac{3}{4}} \cdot x^{\frac{9}{4}} \cdot x^{\frac{5}{4}} = 24 \cdot x^{\frac{3+9+5}{4}} = 24x^{\frac{17}{4}}$$ --- **Part c)** Simplify $\sqrt[3]{a^{2}} \cdot \sqrt[3]{a^{3}}$ $$a^{\frac{2}{3}} \cdot a^{\frac{3}{3}} = a^{\frac{2}{3} + 1} = a^{\frac{5}{3}}$$ --- **Part d)** Simplify $\sqrt[3]{x^{4}} \cdot \sqrt[3]{x^{5}}$ $$x^{\frac{4}{3}} \cdot x^{\frac{5}{3}} = x^{\frac{9}{3}} = x^{3}$$ --- **Part e)** Simplify $\sqrt[3]{\frac{a^{5} b^{7}}{a b^{4}}}$ Simplify inside first: $$\frac{a^{5} b^{7}}{a b^{4}} = a^{5-1} b^{7-4} = a^{4} b^{3}$$ Then cube root: $$\sqrt[3]{a^{4} b^{3}} = a^{\frac{4}{3}} b^{1} = a^{\frac{4}{3}} b$$ --- **Part f)** Simplify partial root $\sqrt[3]{32 d^{13}}$ Rewrite 32 as $2^{5}$: $$\sqrt[3]{2^{5} d^{13}} = 2^{\frac{5}{3}} d^{\frac{13}{3}} = 2^{1 + \frac{2}{3}} d^{4 + \frac{1}{3}} = 2 \cdot 2^{\frac{2}{3}} d^{4} d^{\frac{1}{3}} = 2 d^{4} \cdot \sqrt[3]{4 d}$$ --- **Rational exponents problems:** 1) Simplify $\sqrt[7]{a^{2}} \cdot \sqrt[3]{a^{1}} \cdot a$ Rewrite all as exponents: $$a^{\frac{2}{7}} \cdot a^{\frac{1}{3}} \cdot a^{1} = a^{\frac{2}{7} + \frac{1}{3} + 1} = a^{\frac{2}{7} + \frac{1}{3} + \frac{7}{7}} = a^{\frac{2}{7} + \frac{1}{3} + 1}$$ Find common denominator 21: $$= a^{\frac{6}{21} + \frac{7}{21} + \frac{21}{21}} = a^{\frac{34}{21}}$$ --- 2) Simplify $\sqrt[3]{\frac{x}{\sqrt{x}}}$ Rewrite $\sqrt{x} = x^{\frac{1}{2}}$: $$\frac{x}{x^{\frac{1}{2}}} = x^{1 - \frac{1}{2}} = x^{\frac{1}{2}}$$ Then cube root: $$\sqrt[3]{x^{\frac{1}{2}}} = x^{\frac{1}{6}}$$ --- 3) Simplify $\left(3 \cdot m^{\frac{1}{4}} \cdot \sqrt[3]{m^{-2}} \div \left(\sqrt{m} \cdot \sqrt[6]{m^{5}}\right)\right)^{-2}$ Rewrite all exponents: $$3 \cdot m^{\frac{1}{4}} \cdot m^{-\frac{2}{3}} \div \left(m^{\frac{1}{2}} \cdot m^{\frac{5}{6}}\right) = 3 \cdot m^{\frac{1}{4} - \frac{2}{3}} \cdot m^{-\left(\frac{1}{2} + \frac{5}{6}\right)}$$ Calculate exponents: $$\frac{1}{4} - \frac{2}{3} = \frac{3}{12} - \frac{8}{12} = -\frac{5}{12}$$ $$\frac{1}{2} + \frac{5}{6} = \frac{3}{6} + \frac{5}{6} = \frac{8}{6} = \frac{4}{3}$$ So exponent of denominator is $\frac{4}{3}$, so total exponent: $$-\frac{5}{12} - \frac{4}{3} = -\frac{5}{12} - \frac{16}{12} = -\frac{21}{12} = -\frac{7}{4}$$ Expression inside parentheses: $$3 m^{-\frac{7}{4}}$$ Raise to power $-2$: $$\left(3 m^{-\frac{7}{4}}\right)^{-2} = 3^{-2} m^{\frac{7}{2}} = \frac{1}{9} m^{\frac{7}{2}}$$ --- 4) Simplify $2^{\frac{1}{2}} \cdot 4^{\frac{1}{3}} \cdot 8^{\frac{1}{4}} \cdot 16^{\frac{1}{6}} \cdot 32^{0}$ Rewrite bases as powers of 2: $$2^{\frac{1}{2}} \cdot (2^{2})^{\frac{1}{3}} \cdot (2^{3})^{\frac{1}{4}} \cdot (2^{4})^{\frac{1}{6}} \cdot 1$$ Calculate exponents: $$2^{\frac{1}{2}} \cdot 2^{\frac{2}{3}} \cdot 2^{\frac{3}{4}} \cdot 2^{\frac{4}{6}} = 2^{\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{2}{3}}$$ Sum exponents: $$\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{2}{3} = \frac{6}{12} + \frac{8}{12} + \frac{9}{12} + \frac{8}{12} = \frac{31}{12}$$ Final answer: $$2^{\frac{31}{12}}$$ --- **Summary:** - Use root to exponent conversion. - Combine like bases by adding/subtracting exponents. - Simplify coefficients separately. This approach helps simplify complex root and rational exponent expressions step-by-step.