Simplify each of the following: a) 27^(2)/(3) x 9^(1)/(2) b) 25^-(3)/(2) x ((1)/(5))⁻¹ c) ((27)/(125))^-(2)/(3) x 81^(3)/(4)

$$27 = 3^3$$ $$9 = 3^2$$

$$3^{3 \times \frac{2}{3}} \times 3^{2 \times \frac{1}{2}}$$ $$3^2 \times 3^1$$

$$\boxed{\text{27}}$$

b) $25^{-\frac{3}{2}} \times (\frac{1}{5})^{-1}$

$$25 = 5^2$$ $$\frac{1}{5} = 5^{-1}$$

$$5^{2 \times (-\frac{3}{2})} \times 5^{(-1) \times (-1)}$$ $$5^{-3} \times 5^1$$

$$5^{-3+1}$$ $$5^{-2}$$

$$\boxed{\text{}\frac{1}{25}\text{}}$$

c) $(\frac{27}{125})^{-\frac{2}{3}} \times 81^{\frac{3}{4}}$

$$\frac{27}{125} = \frac{3^3}{5^3} = \left(\frac{3}{5}\right)^3$$ $$81 = 3^4$$

$$\left(\frac{3}{5}\right)^{3 \times (-\frac{2}{3})} \times 3^{4 \times \frac{3}{4}}$$ $$\left(\frac{3}{5}\right)^{-2} \times 3^3$$

$$\frac{5^2}{3^2} \times 27$$ $$\frac{25}{9} \times 27$$

$$25 \times \frac{27}{9}$$ $$25 \times 3$$ $$\boxed{\text{75}}$$

8. Solve each of the following equations

a) $x^2 + 6x + 8 = 0$

We need two numbers that multiply to 8 (P) and add to 6 (S). These numbers are 4 and 2.

$$x^2 + 4x + 2x + 8 = 0$$

$$x+2 = 0 \quad \text{or} \quad x+4 = 0$$ $$x = -2 \quad \text{or} \quad x = -4$$ $$\boxed{\text{x = -2, x = -4}}$$

Questions 3, 5(a), 5(b), 5(c), and 8(b) are incomplete in the image provided. Please send the full questions if you'd like me to solve them.