y=∑an(x-1)n, y'=∑nan(x-1)n-1, y"=∑n(n-1)an(x-1)n-2, if we build the series around x=1.
Expand the first few terms:
y=a0+a1(x-1)+a2(x-1)2+a3(x-1)3+a4(x-1)4+a5(x-1)5+...
y'=a1+2a2(x-1)+3a3(x-1)2+4a4(x-1)3+5a5(x-1)4+6a6(x-1)5+...
y"=2a2+6a3(x-1)+12a4(x-1)2+20a5(x-1)3+30a6(x-1)4+42a7(x-1)5+...
Note that x3-1=(x-1)(x2+x+1), so the DE becomes:
(x3-1)y"+3xy'-y=
(x2+x+1)(2a2+6a3(x-1)2+12a4(x-1)3+20a5(x-1)4+30a6(x-1)5+...
+3x(a1+2a2(x-1)+3a3(x-1)2+4a4(x-1)3+5a5(x-1)4+6a6(x-1)5+...)
-(a0+a1(x-1)+a2(x-1)2+a3(x-1)3+a4(x-1)4+a5(x-1)5+...)=0.
Consider the constant terms:
2a2-a0=0, a2=a0/2, because the DE must be zero for all x.
Consider (x-1) terms:
(x3-1)y"+3xy'-y=0=(x2+x+1)∑n(n-1)an(x-1)n-1+3x∑nan(x-1)n-1-∑an(x-1)n=0,
∑nan(x-1)n-1[(n-1)(x2+x+1)+3x]-∑an(x-1)n=0.
More to follow...