13*. Calculate:
1) \(x^{4}-2 x^{3}+3 x^{2}-2 x+2\) given \(x^{2}-x=3\)
2) \(2 x^{4}+3 x^{2} y^{2}+y^{4}+y^{2} \quad\) given \(x^{2}+y^{2}=1\).
13. 1) \( \mathrm{P} \) is a solution. \( x^{2}\left(x^{2}-x\right)-x\left(x^{2}-x\right)+2\left(x^{2}-x\right)+2= \) \( =3 x^{2}-3 x+2 \cdot 3+2=3\left(x^{2}-x\right)+8=3 \cdot 3+8=17 \);

2) Solution. \( 2 x^{4}+2 x^{2} y^{2}+x^{2} y^{2}+y^{4}+y^{2}=2 x^{2}\left(x^{2}+\right. \) \( \left.+y^{2}\right)+y^{2}\left(x^{2}+y^{2}\right)+y^{2}=2 x^{2} \cdot 1+y^{2} \cdot 1+y^{2}=2\left(x^{2}+y^{2}\right)=2 \).

If students are already familiar with the formulas for abbreviated multiplication, the exercise can be solved by substitution \( x^{2}=1-y^{2} \).