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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.

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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?

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  • $\begingroup$ 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. $\endgroup$
    – H.Linkhorn
    Commented Oct 13, 2018 at 18:53
  • $\begingroup$ By going up to the $x^3$ term within your two expansions you will get $\frac23x^3$. $\endgroup$
    – mrtaurho
    Commented Oct 13, 2018 at 18:55
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    $\begingroup$ How? as i found $\frac{4}{27}x^3$ and $\frac{-4}{27}x^3$ whihc resulted in my having no $x^3$ term $\endgroup$
    – H.Linkhorn
    Commented Oct 13, 2018 at 18:57
  • $\begingroup$ I will add it within my answer. Wait a moment. $\endgroup$
    – mrtaurho
    Commented Oct 13, 2018 at 18:57
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    $\begingroup$ Thank you, i have seen exactly what ive missed now $\endgroup$
    – H.Linkhorn
    Commented Oct 13, 2018 at 19:04

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