Explanation

(a)
$T_{n} = (2n - 1)^{3}$
$= 8n^{3} - 1^{3} - 3.2n.1(2n - 1)$
$= 8n^{3} - 1 - 12n^{2} + 6n$
$= 8n^{3} - 12n^{2} + 6n - 1$
$\therefore S_{n} = \Sigma T_{n}$
$= 8\Sigma n^{3} - 12\Sigma n^{2} + 6\Sigma n - \Sigma 1$
= 8.$\left\lbrack \frac{n(n + 1)}{2} \right\rbrack^{2} - 12.\frac{n(n + 1)(2n + 1)}{6} + 6\frac{n(n + 1)}{2} - n$
$= 2n^{2}(n + 1)^{2} - 2n(n + 1)(2n + 1) + 3n(n + 1) - n$
$= n(n + 1)\left\lbrack 2n(n + 1) - 2(2n + 1) + 3 \right\rbrack - n$
$= n(n + 1)\left\lbrack 2n^{2} + 2n - 4n - 2 + 3 \right\rbrack - n$
$= n(n + 1)\left\lbrack 2n^{2} - 2n + 1 \right\rbrack - n$
$= n(n + 1).2n(n - 1) + n(n + 1) - n$
$= 2n^{2}\left( n^{2} - 1 \right) + n^{2}$
$= n^{2}(2n^{2} - 1)$