1. **Problem statement:** Simplify each expression without using a calculator. 2. **Recall exponent rules:** - $a^0 = 1$ for any $a \neq 0$. - $(a^m)^n = a^{mn}$. - $a^m \cdot a^n = a^{m+n}$. - $\frac{a^m}{a^n} = a^{m-n}$. - $(ab)^n = a^n b^n$. - $(\frac{a}{b})^n = \frac{a^n}{b^n}$. 3. **Simplify each part:** (a) $5a^0 + (5a)^0$ Since $a^0=1$ and $(5a)^0=1$, $$5 \times 1 + 1 = 5 + 1 = 6$$ (b) $(-2x^3 y^2)^3$ Apply power to each factor: $$(-2)^3 (x^3)^3 (y^2)^3 = -8 x^{9} y^{6}$$ (c) $(3^3)^2 \cdot 3^2$ Use $(a^m)^n = a^{mn}$: $$(3^3)^2 = 3^{3 \times 2} = 3^6$$ Multiply: $$3^6 \cdot 3^2 = 3^{6+2} = 3^8$$ (d) $\frac{2x^2 y^4 x y^3}{6x^6 y^6}$ Combine like terms in numerator: $$2 x^{2+1} y^{4+3} = 2 x^3 y^7$$ Divide numerator by denominator: $$\frac{2 x^3 y^7}{6 x^6 y^6} = \frac{\cancel{2} x^{3-6} y^{7-6}}{\cancel{6} 3} = \frac{1}{3} x^{-3} y^{1} = \frac{y}{3 x^3}$$ (e) $\left(\frac{-12 x^3}{15 y^{-2}}\right)^2$ Rewrite denominator: $$15 y^{-2} = 15 \times \frac{1}{y^2} = \frac{15}{y^2}$$ So expression is: $$\left(\frac{-12 x^3}{\frac{15}{y^2}}\right)^2 = \left(-12 x^3 \times \frac{y^2}{15}\right)^2 = \left(\frac{-12}{15} x^3 y^2\right)^2$$ Simplify fraction: $$\frac{-12}{15} = \frac{-4}{5}$$ Square each factor: $$\left(\frac{-4}{5}\right)^2 (x^3)^2 (y^2)^2 = \frac{16}{25} x^{6} y^{4}$$ (f) $\left(\frac{10 y^4}{4 y^6}\right)^{-2}$ Simplify inside parentheses: $$\frac{10}{4} y^{4-6} = \frac{5}{2} y^{-2}$$ Apply negative exponent: $$\left(\frac{5}{2} y^{-2}\right)^{-2} = \left(\frac{5}{2}\right)^{-2} (y^{-2})^{-2} = \left(\frac{2}{5}\right)^2 y^{4} = \frac{4}{25} y^{4}$$ (g) $(-3 a^2 b^4)^2 \cdot 3 (a^2 b)^3$ First part: $$(-3)^2 (a^2)^2 (b^4)^2 = 9 a^{4} b^{8}$$ Second part: $$3 (a^2)^3 b^3 = 3 a^{6} b^{3}$$ Multiply both parts: $$9 a^{4} b^{8} \times 3 a^{6} b^{3} = 27 a^{4+6} b^{8+3} = 27 a^{10} b^{11}$$ (h) $(3 x^3 \times 3 x^3)^2 + (3 x^3 + 3 x^3)^2$ First term inside parentheses: $$3 x^3 \times 3 x^3 = 9 x^{6}$$ Square it: $$(9 x^{6})^2 = 81 x^{12}$$ Second term inside parentheses: $$3 x^3 + 3 x^3 = 6 x^3$$ Square it: $$(6 x^3)^2 = 36 x^{6}$$ Sum both: $$81 x^{12} + 36 x^{6}$$ (i) $\frac{x + y - 1}{x + y}$ Cannot simplify further without values; expression remains: $$\frac{x + y - 1}{x + y}$$ **Final answers:** (a) 6 (b) $-8 x^{9} y^{6}$ (c) $3^{8}$ (d) $\frac{y}{3 x^{3}}$ (e) $\frac{16}{25} x^{6} y^{4}$ (f) $\frac{4}{25} y^{4}$ (g) $27 a^{10} b^{11}$ (h) $81 x^{12} + 36 x^{6}$ (i) $\frac{x + y - 1}{x + y}$