If $a -\dfrac{1}{a}=5$; find $a^2+\dfrac{1}{a^2}-3a + \dfrac{3}{a}$ .

Given: $a - \dfrac{1}{a} = 5$

Squaring both sides, we get

$⇒ \Big(a - \dfrac{1}{a}\Big)^2 = (5)^2\\[1em] ⇒ a^2 + \dfrac{1}{a^2} - 2 \times a \times \dfrac{1}{a} = 25\\[1em] ⇒ a^2 + \dfrac{1}{a^2} - 2 \times \cancel{a} \times \dfrac{1}{\cancel{a}} = 25\\[1em] ⇒ a^2 + \dfrac{1}{a^2} - 2 = 25\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 25 + 2\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 27\\[1em]$

Now,

$a^2+\dfrac{1}{a^2}-3a + \dfrac{3}{a}\\[1em] = \Big(a^2 + \dfrac{1}{a^2}\Big) - 3\Big(a - \dfrac{1}{a}\Big)\\[1em] = 27 - 3 \times 5\\[1em] = 27 - 15\\[1em] = 12$

Hence, $a^2+\dfrac{1}{a^2}-3a + \dfrac{3}{a} = 12$.