本题考查二项式定理求通项的相关问题．将原式子按 $x-y$ 展开，再分别用二项式定理的通项公式求解．因为 $\left(x - y\right){\left(x + y\right)^8}=x\left(x+y\right)^8-y\left(x+y\right)^8$，而 ${x^2}{y^7} = x \cdot \left(x{y^7}\right)= y \cdot \left(x^2{y^6}\right)$，故只需求出 $ \left(x+y\right)^8 $ 的展开式中 $ xy^7 $ 与 $ x^2y^6 $ 的系数即可．
因为 $ \left(x+y\right)^8 $ 的展开式中 $x{y^7}$ 的系数为 ${\mathrm{C}}_8^7$，$x^2{y^6}$ 的系数为 ${\mathrm{C}}_8^6$，所以 $\left(x - y\right){\left(x + y\right)^8}$ 的展开式中 ${x^2}{y^7}$ 的系数为 ${\mathrm{C}}_8^7-{\mathrm{C}}_8^6=-20$．