If $a^2 + \dfrac{1}{a^2} = 23$, find the values of:

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

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

(i) Given,

$a^2 + \dfrac{1}{a^2} = 23$

Using identity,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = a^2 + \dfrac{1}{a^2} + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 23 + 2 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 25 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \sqrt{25} \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \pm 5$

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

(ii) Given,

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

Case 1:

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

We know that,

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

Substituting values we get :

$\Rightarrow 5^3 = a^3 + \dfrac{1}{a^3} + 3 \times 5 \\[1em] \Rightarrow 125 = a^3 + \dfrac{1}{a^3} + 15 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 125 - 15 = 110.$

Case 2:

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

We know that,

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

Substituting values we get :

$\Rightarrow (-5)^3 = a^3 + \dfrac{1}{a^3} + 3 \times (-5) \\[1em] \Rightarrow -125 = a^3 + \dfrac{1}{a^3} - 15 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -125 + 15 = -110.$

Hence, $a^3 + \dfrac{1}{a^3} = \pm 110.$