The system of equations $x^5+y^5=33$ and $x+y=3$ can be solved using polynomial identities. By applying the identity $(x+y)^5=x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4$, the equation simplifies to $15x^2y^2-135xy+210=0$. This leads to two cases for the product $xy$: either $xy=2$ or $xy=7$. The first case yields real solutions $(x=1,y=2)$ and $(x=2,y=1)$, while the second case has no real solutions. Complex solutions are also provided as $x=(3+i\sqrt{19})/2$ and $y=(3-i\sqrt{19})/2$.
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[sp]The complex solutions are:solakis said:[sp]$(x+y)^5=x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4=33+5xy(x^3+y^3)+10x^2y^2(x+y)=
=33+5xy((x+y)^3-3xy(x+y))+30x^2y^2= 33+5xy(27-9xy)+30x^2y^2=243$
or
$15x^2y^2-135xy+210=0$
or
$x^2y^2-9xy+14=0$
And xy=7 or xy=2 impling the following 2 systems of equations :
x+y=3. (A)
xy=2
x+y=3 (B)
xy=7
And (A) gives (x=1,y=2),(x=2,y=1) (B) has no real solutions[/sp]