If $x +\dfrac{1}{x} = 2$, the value of $\Big(x^3 +\dfrac{1}{x^3}\Big) - \Big(x^2 +\dfrac{1}{x^2}\Big)$ is:

1. 0

2. 4

3. 2

4. 6

Using the formula,

[∵ (x + y)³ = x³ + 3xy(x + y) + y³]

And,

[∵ (x + y)² = x² + 2xy + y²]

$\Big(x + \dfrac{1}{x}\Big)^3 = x^3 + 3 \times x \times \dfrac{1}{x}\Big(x + \dfrac{1}{x}\Big) + \Big(\dfrac{1}{x}\Big)^3\\[1em] ⇒ \Big(x + \dfrac{1}{x}\Big)^3 = x^3 + 3\Big(x + \dfrac{1}{x}\Big) + \dfrac{1}{x^3}$

Putting $x +\dfrac{1}{x} = 2$

$2^3 = x^3 + 3 \times 2 + \dfrac{1}{x^3}\\[1em] ⇒ 8 = x^3 + 6 + \dfrac{1}{x^3}\\[1em] ⇒ x^3 + \dfrac{1}{x^3} = 8 - 6\\[1em] ⇒ x^3 + \dfrac{1}{x^3} = 2……………(1)$

And ,

$\Big(x + \dfrac{1}{x}\Big)^2 = x^2 + 2 \times x \times \dfrac{1}{x} + \Big(\dfrac{1}{x}\Big)^2\\[1em] ⇒ \Big(x + \dfrac{1}{x}\Big)^2 = x^2 + 2 + \dfrac{1}{x^2}$

Putting the value $x +\dfrac{1}{x} = 2$,we get

$2^2 = x^2 + 2 + \dfrac{1}{x^2}\\[1em] ⇒ 4 = x^2 + 2 + \dfrac{1}{x^2}\\[1em] ⇒ 4 - 2 = x^2 + \dfrac{1}{x^2}\\[1em] ⇒ 2 = x^2 + \dfrac{1}{x^2} ……………(2)$

Now, using equation (1) and (2),

$\Big(x^3 +\dfrac{1}{x^3}\Big) - \Big(x^2 +\dfrac{1}{x^2}\Big)\\[1em] = 2 - 2\\[1em] = 0$

Hence, option 1 is the correct option.