10. If $x, y$ are real numbers, and $x^{2}+x y+y^{2}=3$, then the maximum and minimum values of $x^{2}-x y+y^{2}$ are $\qquad$.
$10.9,1$.
Let $x^{2}-x y+y^{2}=m$, then
$$
x+y= \pm \sqrt{\frac{9-m}{2}}, x y=\frac{3-m}{2} \text {. }
$$

Therefore, $x, y$ are the two real roots of the equation $t^{2} \pm \sqrt{\frac{9-m}{2}} t+\frac{3-m}{2}=0$. By $\Delta \geqslant 0$, we get $m \geqslant 1$; by $\frac{9-m}{2} \geqslant 0$, we get $m \leqslant 9$.