If $a^2 + \dfrac{1}{a^2} = 27$, find the values of :

(i) $\Big(a - \dfrac{1}{a}\Big)$

(ii) $\Big(a^3 - \dfrac{1}{a^3}\Big)$

(i) Given,

$a^2 + \dfrac{1}{a^2} = 27$

Using identity,

$\Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 27 - 2 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = 25 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = \sqrt{25} \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = \pm 5$

Hence, $\Big(a - \dfrac{1}{a}\Big)^2 = \pm 5.$

(ii) Given,

$\Big(a + \dfrac{1}{a}\Big) = \pm 5.$

Case 1:

$\Big(a + \dfrac{1}{a}\Big) = +5.$

We know that,

$\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = \Big(a - \dfrac{1}{a}\Big)^3 + 3 \times a \times \dfrac{1}{a} \Big(a - \dfrac{1}{a}\Big)$

Substituting values we get :

$\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (5)^3 + 3(5) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (125) + (15) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = 140.$

Case 2:

$\Big(a + \dfrac{1}{a}\Big) = -5.$

We know that,

$\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = \Big(a - \dfrac{1}{a}\Big)^3 + 3 \times a \times \dfrac{1}{a} \Big(a - \dfrac{1}{a}\Big)$

Substituting values we get :

$\Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (-5)^3 + 3(-5) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = (-125) + (-15) \\[1em] \Rightarrow \Big(a^3 - \dfrac{1}{a^3}\Big) = -140.$

Hence, $a^3 - \dfrac{1}{a^3} = \pm 140$