If $S_{n}$ denotes the sum of the first $n$ terms of an $AP$,prove that $S_{12} = 3(S_{8} - S_{4})$.

The sum of the first $n$ terms of an $AP$ is given by the formula: $S_{n} = \frac{n}{2}[2a + (n - 1)d]$ ... $(i)$
Calculating $S_{8}$:
$S_{8} = \frac{8}{2}[2a + (8 - 1)d] = 4(2a + 7d) = 8a + 28d$
Calculating $S_{4}$:
$S_{4} = \frac{4}{2}[2a + (4 - 1)d] = 2(2a + 3d) = 4a + 6d$
Now,calculating the difference $(S_{8} - S_{4})$:
$S_{8} - S_{4} = (8a + 28d) - (4a + 6d) = 4a + 22d$ ... $(ii)$
Calculating $S_{12}$:
$S_{12} = \frac{12}{2}[2a + (12 - 1)d] = 6(2a + 11d) = 12a + 66d$
From equation $(ii)$,we can see that $3(S_{8} - S_{4}) = 3(4a + 22d) = 12a + 66d$.
Since $S_{12} = 12a + 66d$ and $3(S_{8} - S_{4}) = 12a + 66d$,it is proved that $S_{12} = 3(S_{8} - S_{4})$.