1. **Problem:** Calculate the expressions: i) $32^{3/5}$ ii) $20^7 \div 20^5$ iii) $27^{-1/3}$ 2. **Work:** i) Recall that $32 = 2^5$, so: $$32^{3/5} = (2^5)^{3/5} = 2^{5 \times \frac{3}{5}} = 2^3 = 8$$ ii) Use the quotient rule for exponents: $$20^7 \div 20^5 = 20^{7-5} = 20^2 = 400$$ iii) Recall that $27 = 3^3$, so: $$27^{-1/3} = \frac{1}{27^{1/3}} = \frac{1}{(3^3)^{1/3}} = \frac{1}{3^{3 \times \frac{1}{3}}} = \frac{1}{3^1} = \frac{1}{3}$$ --- 3. **Problem:** Given equations: $$3^{x-y} = 27$$ $$3^{x+y} = 243$$ Find values of $x$ and $y$. 4. **Work:** Express constants as powers of 3: $$27 = 3^3$$ $$243 = 3^5$$ Rewrite: $$3^{x-y} = 3^3 \implies x - y = 3$$ $$3^{x+y} = 3^5 \implies x + y = 5$$ Add the equations: $$ (x - y) + (x + y) = 3 + 5 \implies 2x = 8 \implies x = 4$$ Substitute back to find $y$: $$x - y = 3 \implies 4 - y = 3 \implies y = 1$$ 5. **Answer:** $$\boxed{8, 400, \frac{1}{3}, x = 4, y = 1}$$