Solve :

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

Let $x - \dfrac{1}{x} = a$ ……..(i)

Squaring, both sides we get,

$\Rightarrow x^2 + \dfrac{1}{x^2} - 2 = a^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = a^2 + 2 …….(ii)$

Substituting the values from equations (i) and (ii) we get,

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

⇒ a² + 2 - 3a - 2 = 0

⇒ a² - 3a = 0

⇒ a(a - 3) = 0

⇒ a = 0 or a - 3 = 0

⇒ a = 0 or a = 3.

Considering a = 0 we get,

$\Rightarrow x - \dfrac{1}{x} = 0 \\[1em] \Rightarrow \dfrac{x^2 - 1}{x} = 0 \\[1em] \Rightarrow x^2 - 1 = 0 \\[1em] \Rightarrow (x - 1)(x + 1) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x + 1 = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = -1.$

Considering a = 3 we get,

$\Rightarrow x - \dfrac{1}{x} = 3 \\[1em] \Rightarrow \dfrac{x^2 - 1}{x} = 3 \\[1em] \Rightarrow x^2 - 1 = 3x \\[1em] \Rightarrow x^2 - 3x - 1 = 0$

Comparing x² - 3x - 1 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -3 and c = -1.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} \\[1em] = \dfrac{3 \pm \sqrt{9 + 4}}{2} \\[1em] = \dfrac{3 \pm \sqrt{13}}{2}.$

Hence, x = 1, -1, $\dfrac{3 \pm \sqrt{13}}{2}$.