Find the binomial expansion of the function up to the first 3 non-zero terms
$$\sqrt[3]\frac{1+2x}{1-x}$$
The function can be broken into $(1+2x)^{\frac{1}{3}}$ and $(1-x)^{\frac{-1}{3}}$
where
$(1+2x)^{\frac{1}{3}} = 1+\frac{2}{3}x-\frac{4}{9}x^2$ and $(1-x)^{\frac{-1}{3}}=1+\frac{1}{3}x+\frac{2}{9}x^2$
however when i multpily out the two expansions i get an $x^3$ term of $0x^3$ when the answer says $\frac{2}{3}x^3$, any help would be appreciated.
Best Answer
HINT
Hence you only got $x^2$ terms within our expansion but the product of $1$ with $x^3$ also gives you an cubic term you need to add the $x^3$ terms within both expansions in order to get the right expansion.
$$\begin{align} (1+2x)^{\frac13}&=1+\frac23x-\frac49x^2+\frac{40}{81}x^3-\frac{160}{243}x^4+\cdots\\(1-x)^{-\frac13}&=1+\frac13x+\frac29x^2+\frac{14}{81}x^3+\frac{35}{243}x^4+\cdots \end{align}$$
By multiplying these two polynomials we will get every possible term from $x^0$ up to $x^8$. But note by only considering the expansion up to $x^2$ we will miss the $x^3$ term constructed out of $x^0$ and $x^3$. Similiar will happen with the $x^4$ constructed from $x$ and $x^3$. You missed these cases.
For example for $x^3$ we will get as coefficients
$$1\cdot\frac{14}{81}+1\cdot\frac{41}{81}+\frac23\cdot\left(\frac{2}{9}\right)+\frac13\cdot\left(-\frac{4}{9}\right)=\frac23+0$$
you only thought about the last two terms and missed the first two. Something similiar can be done with $x^4$ terms and so on. Everything clear now?