Find the coefficient of $x^8$ in the  expansion of ${\left(x^2-\frac{1}{x}\right)}^{10}$

We know, 

General term in $(x+a{)}^n$ given by

$T_{r+1}=C(n,r)x^{n-r}{a}^r$

$\therefore$ General term in ${\left(x^2-\frac{1}{x}\right)}^{10}$

$\begin{array}{r}{T}_{r+1}=C(10,r){\left(x^2\right)}^{10-r}{\left(-\frac{1}{x}\right)}^r\\ \Rightarrow T_{r+1}=(-1{)}^{r}C(10,r)x^{20-2r}{x}^{-r}\end{array}$

The word with a coefficient of$x^8$ must now be found.

$$\begin{gathered} \therefore 20-2r-r=8 \\ \Rightarrow r=4 \end{gathered}$$

So, ${\mathrm{T}}^{5 }$will have coefficient $x^8$

$\begin{array}{l}\therefore T_5=(-1{)}^{4}C(10,4)x^8\\ \Rightarrow T_5=\frac{10\cdot 9\cdot 8\cdot 7}{4\cdot 3\cdot 2\cdot 1}{x}^8\\ \Rightarrow T_5=210x^8\end{array}$

So, the coefficient of  $x^8$ in the given expansion will be $210.$