Question

Find the $6^{th}$ term in the expansion of ${(x^3-\frac{1}{x^2})}^{10}$

Answer 1

The given term is ${\left(x^3-\frac{1}{x^2}\right)}^{10}$.

The general term of expansion of ${\left(a+b\right)}^n$ is,

$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

The sixth term of the expansion of ${\left(x^3-\frac{1}{x^2}\right)}^{10}$ is,

$T_{5+1}={}^{10}{C}_5{\left(x^3\right)}^{10-5}{\left(-\frac{1}{x^2}\right)}^5$

$=-\frac{10!}{5!\left(10-5\right)!}{\left(x^3\right)}^5\left(\frac{1}{{\left(x^2\right)}^5}\right)$

$=-\frac{10!}{5!5!}\left(x^{15}\right)\left(\frac{1}{x^{10}}\right)$

$=-252x^{15-10}$

$=-252x^5$