Find : $a^3 +\dfrac{1}{a^3}$, if $a +\dfrac{1}{a}= 5$.

Using the formula,

[∵ (x + y)³ = x³ + 3xy(x + y) + y³]

So,

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

Putting $a +\dfrac{1}{a} = 5$

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

Hence, the value of $a^3 + \dfrac{1}{a^3}$ is 110.