217. a) x³/3 + x² - 3x + 2 = 0;

Let P(x) = x³/3 + x² - 3x + 2. We look for rational roots of the form p/q, where p divides 6 and q divides 1. Possible rational roots are ±1, ±2, ±3, ±6. P(3) = 27/3 + 9 - 9 + 2 = 9 + 2 = 11. P(-3) = -27/3 + 9 + 9 + 2 = -9 + 20 = 11. P(1) = 1/3 + 1 - 3 + 2 = 1/3. P(-1) = -1/3 + 1 + 3 + 2 = 5 - 1/3 = 14/3. P(2) = 8/3 + 4 - 6 + 2 = 8/3. P(-2) = -8/3 + 4 + 6 + 2 = 12 - 8/3 = 28/3. P(6) = 216/3 + 36 - 18 + 2 = 72 + 20 = 92. P(-6) = -216/3 + 36 + 18 + 2 = -72 + 56 = -16. Let's check integer roots for x³ + 3x² - 9x + 6 = 0. P(1) = 1+3-9+6 = 1. P(-1) = -1+3+9+6 = 17. P(2) = 8+12-18+6 = 8. P(-2) = -8+12+18+6 = 28. P(3) = 27+27-27+6 = 33. P(-3) = -27+27+27+6 = 33. P(-4) = -64+48+36+6 = 26. P(-5) = -125+75+45+6 = 1. P(-6) = -216+108+54+6 = -48. Let's consider the original equation: x³/3 + x² - 3x + 2 = 0. Multiply by 3: x³ + 3x² - 9x + 6 = 0. Let f(x) = x³ + 3x² - 9x + 6. f'(x) = 3x² + 6x - 9 = 3(x² + 2x - 3) = 3(x+3)(x-1). Critical points at x = -3 and x = 1. f(-3) = (-3)³ + 3(-3)² - 9(-3) + 6 = -27 + 27 + 27 + 6 = 33. f(1) = 1³ + 3(1)² - 9(1) + 6 = 1 + 3 - 9 + 6 = 1. Since the local maximum f(-3) = 33 > 0 and the local minimum f(1) = 1 > 0, and lim(x→-∞) f(x) = -∞, there is one real root less than -3.