Question

Find the $(r+1{)}^{th}$ term in the following expansion :
$\sqrt[3]{3(a^3-x^3{)}^2}$

Answer 1

The general term i.e. $(r+1{)}^{th}$ term in the expansion of $(1+x{)}^n$ is given by

$T_{r+1}=\frac{n(n-1)(n-2)....(n-r+1)}{r!}{x}^r$

Now,

$\sqrt[3]{3(a^3-x^3{)}^2}=a^2{(1-\frac{x^3}{a^3})}^{\frac{2}{3}}$

$T_{r+1}=a^2\left[\frac{\frac{2}{3}(\frac{2}{3}-1)(\frac{2}{3}-2)(\frac{2}{3}-3)..........(\frac{2}{3}-r+1)}{r!}{(-\frac{x^3}{a^3})}^r\right]$

$=a^2\left[\frac{2.(-1)(-4)(-7)........(5-3r)}{3^r.r!.a^{3r}}\right(-1{)}^r{x}^{3r}]$

$=(-1{)}^r(-1{)}^{r-1}\frac{\mathrm{2.1.4.7.......}(3r-5)}{3^r.r!.a^{3r-2}}{x}^{3r}$

$=-\frac{\mathrm{2.1.4.7.......}(3r-5)}{3^r.r!.a^{3r-2}}{x}^{3r}$