If $\dfrac{a}{b} + \dfrac{b}{a} = 1$ (a, b ≠ 0), then a³ + b³ =

1. 0

2. 1

3. 2

4. 8

Given,

$\Rightarrow \dfrac{a}{b} + \dfrac{b}{a} = 1 \\[1em] \Rightarrow \dfrac{a^2 + b^2}{ab} = 1 \\[1em] \Rightarrow a^2 + b^2 = ab \\[1em] \Rightarrow a^2 + b^2 - ab = 0.$

We know that,

⇒ a³ + b³ = (a + b)(a² - ab + b²)

⇒ a³ + b³ = (a + b)(0)

⇒ a³ + b³ = 0.

Hence, Option 1 is the correct option.