Find $x,y \in \mathbb C $ such that $x^5+y^5=275, x+y=5$. My attempt:
Let $x, y$ be roots of
$$t^2-5t+p=0$$
I got
$$p=\frac{5 \pm \sqrt{-431}}{2}$$
using Vieta's relations.
Now I just calculated x and y using the quadratic formula in $t^2-5t+p=0$.
But I got two very hideous complex values of x and y. So I think my answer may be wrong.
The values I got are
$$x= \frac{5+\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}$$ or
$$x= \frac{5+\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}$$
Please post your own solutions too if you have any different solutions.

$$x^3+y^3=(x+y)^3-3xy(x+y)=5^3-3xy(5)$$
$$x^2+y^2=(x+y)^2-2xy=5^2-2xy$$
$$(x^3+y^3)(x^2+y^2)=x^5+y^5+x^2y^2(x+y)$$
$$\iff(125-15xy)(25-2xy)=275+5x^2y^2$$
$$\iff(xy)^2-25xy+114=0$$
For the rest, follow How to solve for $x$ in $\sqrt[4]{x+27}+\sqrt[4]{55-x}=4$?
