3. Real numbers $a$ and $b$ are such that $a^{5}+b^{5}=3, a^{15}+b^{15}=9$. Find the value of the expression $a^{10}+b^{10}$.
Answer: 5.
Solution. Let $x=a^{5}, y=b^{5}$. Then $x+y=3, x^{3}+y^{3}=9 ;(x+y)\left(x^{2}-x y+y^{2}\right)=9$; $3\left(x^{2}-x y+y^{2}\right)=9 ; x^{2}-x y+y^{2}=3 ; x^{2}+2 x y+y^{2}=9 ; 3 x y=6 ; x y=2 ; x^{2}+y^{2}=9-2 x y=9-$ $2 \cdot 2=5 ; a^{10}+b^{10}=x^{2}+y^{2}=5$

Criteria. Simplification using substitution was done, but the solution is not complete: 1 point.

The value $x y=2$ was found, but the solution is not complete: 3 points.