If $a + \dfrac{1}{a} = p$, prove that $a^3 + \dfrac{1}{a^3} = p(p^2 - 3)$.

We know that,

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

Substituting values we get,

$a^3 + \dfrac{1}{a^3} = p^3 - 3p \\[1em] = p(p^2 - 3).$

Hence, proved that $a^3 + \dfrac{1}{a^3} = p(p^2 - 3)$.