1. Let $x, y$ be numbers, $x^{2}+x y+y^{2}=3$. Then, the value range of $x^{2}-x y+y^{2}$ is $\qquad$
ii. $1.1 \leqslant x^{2}-x y+y^{2} \leqslant 9$.
Let $x^{2}-x y+y^{2}=k$.
Also, $x^{2}+x y+y^{2}=3$.
From ((2) - (1)), we get $x y=\frac{3-k}{2}$.
From (2), we have $(x+y)^{2}=3+x y=\frac{9-k}{2}$.
Since $(x+y)^{2} \geqslant 0$,
Therefore, $k \leqslant 9$. At this point, $x+y= \pm \sqrt{\frac{9-k}{2}}$.
Therefore, $x$ and $y$ are the roots of the equation $t^{2} \mp \sqrt{\frac{9-k}{2}} t+\frac{3-k}{2}=0$.
Therefore, $\Delta=\left(\mp \sqrt{\frac{9-k}{2}}\right)^{2}-4 \cdot \frac{3-k}{2} \geqslant 0$.
Solving, we get $k \geqslant 1$.
Hence, $1 \leqslant x^{2}-x y+y^{2} \leqslant 9$.