Given: $S_n=\sum n$, $S_2=\sum n^2$ and $S_3=\sum n^3$
$\Rightarrow\:S_1=\sum n=\large\frac{n(n+1)}{2}$
$\:\:\:S_2=\sum n^2=\large\frac{n(n+1)(2n+1)}{6}$$=\large\frac{n(n+1)}{2}\times\frac{2n+1}{3}$
$\Rightarrow\:S_2=S_1.\large\frac{2n+1}{3}$
$\Rightarrow\:3S_2=S_1(2n+1)$............(i)
$S_3=\sum n^3=\large\frac{n^2(n+1)^2}{4}=\bigg[\large\frac{n(n+1)}{2}\bigg]^2$$=S_1^2$.......(ii)
From (i) we get $(3S_2)^2=9S_2^2=S_1^2(2n+1)^2$..........(iii)
Step 2
$S_3(1+8S_1)=S_1^2(1+8S_1)$ (since $S_3=S_1^2.)$
$\qquad\:=S_1^2(1+8\large\frac{n(n+1)}{2})$
$\qquad\:=S_1^2(1+4n^2+4n)$
$\qquad\:=S_1^2(1+2n)^2$
$\qquad\:=9S_2^2$ (From (iii))
$\therefore\: S_3(1+8S_1)=9S_2^2$
Hence proved.