If $S_{1}, S_{2}, S_{3}$ are the sum of first $n$ natural numbers,their squares,and their cubes,respectively,show that $9 S_{2}^{2} = S_{3}(1 + 8 S_{1})$.

(N/A) Given that:
$S_{1} = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
$S_{2} = \sum_{k=1}^{n} k^{2} = \frac{n(n+1)(2n+1)}{6}$
$S_{3} = \sum_{k=1}^{n} k^{3} = \frac{n^{2}(n+1)^{2}}{4}$
Consider the right-hand side $(RHS)$: $S_{3}(1 + 8 S_{1})$
$= \frac{n^{2}(n+1)^{2}}{4} \left(1 + 8 \cdot \frac{n(n+1)}{2}\right)$
$= \frac{n^{2}(n+1)^{2}}{4} (1 + 4n(n+1))$
$= \frac{n^{2}(n+1)^{2}}{4} (1 + 4n^{2} + 4n)$
$= \frac{n^{2}(n+1)^{2}}{4} (2n+1)^{2}$
$= \frac{[n(n+1)(2n+1)]^{2}}{4} \quad \dots (1)$
Now consider the left-hand side $(LHS)$: $9 S_{2}^{2}$
$= 9 \left( \frac{n(n+1)(2n+1)}{6} \right)^{2}$
$= 9 \cdot \frac{[n(n+1)(2n+1)]^{2}}{36}$
$= \frac{[n(n+1)(2n+1)]^{2}}{4} \quad \dots (2)$
From $(1)$ and $(2)$,we have $9 S_{2}^{2} = S_{3}(1 + 8 S_{1})$.