Finding the Maclaurin polynomial of order 6 of: $f(x)=x\ln(1+x^{3})\ln(1-x^{2})$ Find the Maclaurin polynomial of order 6 of:
$$f(x)=x\ln(1+x^{3})\ln(1-x^{2})$$
The result I get doesn't make sense.
my try:
because:  $\ln(1+x)=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n+1}}{n+1}$
$$\ln(1+x^{3})=x^{3}-\frac{x^{6}}{2}+\frac{x^{9}}{3}+x^{9}\varepsilon_{1}(x^{3}) = x^{3}+x^{5}\varepsilon_{2}(x)$$
$$\ln(1-x^{2})=-x^{2}-\frac{x^{4}}{2}-\frac{x^{6}}{6}-x^{6}\varepsilon_{1}(-x^{2}) = -x^{2}-\frac{x^{4}}{2}+x^{5}\varepsilon_{3}(x)$$
$f(x)=x\ln(1+x^{3})(1-x^{2})=x(x^{3}+x^{5}\varepsilon_{2}(x))(-x^{2}-\frac{x^{4}}{2}+x^{5}\varepsilon_{3}(x))$
$=(x^{4}+x^{6}\varepsilon_{2}(x))(-x^{2}-\frac{x^{4}}{2}+x^{5}\varepsilon_{3}(x))=-x^{6}+x^{6}\varepsilon_{4}(x)\Longrightarrow P_{6,0}(x)=-x^{6}$
But this doesn't make sense at all, I guess I wrote nonsense :(
Would appreciate any help :)

after writing out the first terms of the maclaurin expansion of $\ln(1+x^3)$ and $\ln(1-x^2)$ , observe that the only way to get an $x^6$ term is to multiply $x$ and the 2 first terms $x^3$ and $-x^2$. thus the order 6 maclaurin polynomial has only one term $-x^6$.
the maclaurin expansion does not have an $x^7$ term as there are no positive integer solutions $(a,b)$ for the eqn $1+2a+3b=7$.
