# Find the binomial expansion of the function up to the first 3 non-zero terms $\sqrt[3]\frac{1+2x}{1-x}$

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.

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?

• i found the product of the two expansions to find a expansion up to a $x^4$ term, however the $x^3$ term isnt what i wanted and so i wondered if i was doing it right. Commented Oct 13, 2018 at 18:53
• By going up to the $x^3$ term within your two expansions you will get $\frac23x^3$. Commented Oct 13, 2018 at 18:55
• How? as i found $\frac{4}{27}x^3$ and $\frac{-4}{27}x^3$ whihc resulted in my having no $x^3$ term Commented Oct 13, 2018 at 18:57
• I will add it within my answer. Wait a moment. Commented Oct 13, 2018 at 18:57
• Thank you, i have seen exactly what ive missed now Commented Oct 13, 2018 at 19:04