5. Find: 1) \(x^{6}+3 x^{2} y^{2}+y^{6}\), if \(x^{2}+y^{2}=1\); 2) \(x^{4}+\) \(+x^{2} y^{2}+y^{4}\), if \(x^{2}+y^{2}=a\) and \(x y=b\).
5. 1) \(\mathrm{P}\) is the solution. First method. The given polynomial is equal to

\[
\left(x^{2}+y^{2}\right)\left(x^{4}-x^{2} y^{2}+y^{4}\right)+3 x^{2} y^{2}=\ldots=\left(x^{2}+y^{2}\right)^{2}=1
\]

Second method. Since \(y^{2}=1-x^{2}\), then, performing the substitution, we get that the given polynomial is identically equal to 1.