If α and β are the greatest divisors of n (n^ 2 - 1) and 2n (n^ 2 +...

For $n \in N ⇒ n = 1, 2, 3 \ldots$

The values of $n \left(n^{2} - 1\right)$ are

$n = 1 ⇒ 1\left(1 - 1\right) = 0$

$n = 2 ⇒ 2\left(4 - 1\right) = 2\left(3\right) = 6$

$n = 3 ⇒ 3 \left(9 - 1\right) = 24 \ldots$

The values of $2n\left(n^{2} + 2\right)$ are,

$n = 1 ⇒ 2\left(1 + 2\right) = 6$

$n = 2 ⇒ 2\left(2\right)\left({2}^{2} +2\right) = 4\left(6\right) = 24$

n = 3 ⇒ 2\left(3\right)\left({3}^{2} +2\right) = 6\left(11\right) = 66 \ldots$</p><p>$∴$~Greatest divisors of~\alpha and \beta~are both~$6$,</p><p>$∴ \alpha\beta = \left(6\right)\left(6\right) = 36