I don't know why I found 2 different answers. When using the product of the series then the binomial identity
$$\sin^2(x) \cos(x)=\sin(2x)\sin(x/2) $$ I found $$\sin^2(x) \cos(x)=\sum x^2\frac{(-4)^n (9^{n+1}-1)x^{2n}}{(2n+2)!4^{n+1}}.$$ By converting $\sin^2(x)\cos(x)$ into $$\sin^2(x)\cos(x)=(1/4)\cos(x) -(1/4)\cos(3x)$$ I got $$\sin^2(x) \cos(x)=\frac{1\sum{(-1)^n x^{2n}(1-9)^n}}{4\cdot(2n)!}.$$ Which of them is true?
Do you mean $$\sin^2 x\cos x?$$ This is $$\frac12\sin x\sin 2x.$$ Using the identity, $$\sin A\sin B=\frac{\cos(A-B)-\cos(A+B)}2$$ we get $$\sin^2 x\cos x=\frac{\cos x-\cos3x}4.$$ Now use $\cos y=\sum_n (-1)^ny^{2n}/(2n)!$.
Your first power series is correct. The second attempt has promise. You have correctly given $\sin(x)^2cos(x)=(1/4)\cos(x)-(1/4)\cos(3x)$ which should give the same answer but you did not expand the two power series correctly. In particular, $(1-9)^n$ should be $(1^n-9^n)$.
