$$(1-x^3)^{n} = \sum_{r=0}^n a_{r}x^{r}(1-x)^{3n-2r}$$ where n belongs to the set of natural numbers.

Clearly, it should be solved using Binomial theorem. But i don't seem to find out how. Please help me by providing a solution and do mention the steps. Thanks.



For $x\ne1,$ $$\dfrac{1-x^3}{(1-x)^3}=\dfrac{1+x+x^2}{(1-x)^2}=\dfrac{(1-x)^2+3x}{(1-x)^2}=1+3\cdot\dfrac x{(1-x)^2}$$

$$\implies\left(\dfrac{1-x^3}{(1-x)^3}\right)^n=\left(1+3\cdot\dfrac x{(1-x)^2}\right)^n$$

Now look at the Binomial series

  • $\begingroup$ I can see it now. So the answer comes out as C(n, r). 3^r? $\endgroup$ – Phill2 Feb 23 '16 at 16:07
  • $\begingroup$ @Phill2, Rightly so $\endgroup$ – lab bhattacharjee Feb 23 '16 at 16:08

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