Finding the Maclaurin polynomial of order 6 of: $f(x)=x\ln(1+x^{3})\ln(1-x^{2})$
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 $1+2a+3b=7$.