If $x = \dfrac{a + b + c}{3}$, then (x - a)³ + (x - b)³ + (x - c)³ can be factorized as:

1. (x - a)(x - b)(x - c)

2. $\dfrac{(x - a)(x - b)(x - c)}{3}$

3. 3(x - a)(x - b)(x - c)

4. none of these

Given,

$x = \dfrac{a + b + c}{3}$

Let, p = x - a, q = x - b, r = x - c

Adding,

⇒ p + q + r = (x - a) + (x - b) + (x - c)

⇒ p + q + r = 3x - a - b - c

⇒ p + q + r = 3x - (a + b + c)

⇒ p + q + r = 3 $\Big(\dfrac{a + b + c}{3}\Big)$ - (a + b + c)

⇒ p + q + r = (a + b + c) - (a + b + c)

⇒ p + q + r = 0

If p + q + r = 0, we use identity,

p³ + q³ + r³ = 3pqr

⇒ (x - a)³ + (x - b)³ + (x - c)³ = 3(x - a)(x - b)(x - c).

Hence, option 3 is correct option.