Find the values of a and b, if (x - 1) and (x + 2) are both factors of (x³ + ax² + bx - 6).

Let f(x) = x³ + ax² + bx - 6

Since (x − 1) and (x + 2) are factors, by the factor theorem, f(1) = 0 and f(−2) = 0.

⇒ f(1) = 0

⇒ (1)³ + a(1)² + b(1) - 6 = 0

⇒ 1 + a + b - 6 = 0

⇒ a + b - 5 = 0

⇒ a + b = 5 ….(1)

⇒ f(-2) = 0

⇒ (-2)³ + a(-2)² + b(-2) - 6 = 0

⇒ -8 + 4a - 2b - 6 = 0

⇒ 4a - 2b = 14

⇒ 2(2a - b) = 14

⇒ 2a - b = $\dfrac{14}{2}$

⇒ 2a - b = 7 ….(2)

Adding equations (1) and (2), we get:

⇒ a + b + 2a - b = 5 + 7

⇒ 3a = 12

⇒ a = $\dfrac{12}{3}$

⇒ a = 4.

Substituting value of a in equation (1), we get :

⇒ 4 + b = 5

⇒ b = 5 - 4

⇒ b = 1.

Hence, the value of a = 4 and b = 1.