Solve $$\begin{cases} (x_{1}+x_{2}+x_{3})^5=3x_{4}\\ (x_{2}+x_{3}+x_{4})^5=3x_{5}\\ (x_{3}+x_{4}+x_{5})^5=3x_{1}\\ (x_{4}+x_{5}+x_{1})^5=3x_{2}\\ (x_{5}+x_{1}+x_{2})^5=3x_{3} \end{cases}$$ all real roots, it seems not easy.
My try:let $$u_{1}=x_{1}+x_{2}+x_{3},u_{2}=x_{2}+x_{3}+x_{4},u_{3}=x_{3}+x_{4}+x_{5},u_{4}=x_{4}+x_{5}+x_{1},u_{5}=x_{5}+x_{1}+x_{2}$$
then $$\begin{cases} u^5_{1}=-u_{1}+2u_{2}-u_{3}+2u_{4}-u_{5}\cdots\cdots (1)\\ u^5_{2}=-u_{1}-u_{2}+2u_{3}-u_{4}+2u_{5}\cdots\cdots (2)\\ u^5_{3}=2u_{1}-u_{2}-u_{3}+2u_{4}-u_{5}\cdots\cdots (3)\\ u^5_{4}=-u_{1}+2u_{2}-u_{3}-u_{4}+2u_{5}\cdots\cdots (4)\\ u^5_{5}=2u_{1}-u_{2}+2u_{3}-u_{4}-u_{5}\cdots\cdots(5) \end{cases}$$
I think this is nice problem, why someone downvote it?
I am assuming that you wish to find all real solutions to your system of five equations.
Claim 1: If at least three of the $x_i$ are $\ge\frac13$, then they all are.
Case 1: Suppose there are at least three adjacent $x_i$ (that is, with indices cyclic, as per the equations) which are $\ge\frac13$. For example, $x_1,x_2,x_3\ge\frac13$. Then $$\begin{array}\\ 1\le x_1+x_2+x_3\\ 1\le (x_1+x_2+x_3)^5=3x_4\\ \end{array}$$ So $x_4\ge\frac13$. Then similarly, $x_5\ge\frac13$.
Case 2: Assume otherwise. Then there are two non-adjacent $x_i$ with values $\lt\frac13$. For example, say $x_4,x_1\lt\frac13$, and all others $\ge\frac13$. Then $$\begin{array}\\ (x_3+x_4+x_5)^5=3x_1\lt 1\\ (x_4+x_5+x_1)^5=3x_2\ge 1\\ \end{array}$$ So $$\begin{array}\\ -x_3-x_4-x_5\gt -1\\ +x_4+x_5+x_1\ge 1\\ \end{array}$$ Adding these two inequalities gives $-x_3+x_1\gt 0$, or $x_1\gt x_3$, a contradiction.
Claim 2: If at least three of the $x_i$ are $\le\frac13$, they all are. Proof is exactly the same as Claim 1, except reverse the inequalities.
Claim 3: If at least three of the $x_i$ are $\ge 0$, they all are. Proof is exactly the same as Claim 1, replacing $\frac13$ and $1$ with $0$.
Claim 4: If all the $x_i\ge 0$, and at least one is $0$, they all are. Suppose $x_4 = 0$, and one or more of $x_1,x_2,x_3 \gt 0$. Then $x_1+x_2+x_3 \gt 0$, so $(x_1+x_2+x_3)^5 = 3x_4 \gt 0$, a contradiction. So $x_1=x_2=x_3 = 0$, and likewise $x_5=0$.
Claim 5: If all the $x_i\le \frac13$, and at least one is $\frac13$, they all are. Suppose $x_4 = \frac13$, and one or more of $x_1,x_2,x_3 \lt \frac13$. Then $x_1+x_2+x_3 \lt 1$, so $(x_1+x_2+x_3)^5 = 3x_4 \lt 1$, a contradiction. So $x_1=x_2=x_3 = \frac13$, and likewise $x_5=\frac13$.
Claim 6: If at least one $x_i$ is positive, the maximum of the $x_i$ is $\ge\frac13$. Let $x_4$ say be equal to the maximum. Then
$$\begin{array}\\ x_1+x_2+x_3\le 3x_4\\ (x_1+x_2+x_3)^5 \le (3x_4)^5\\ 3x_4\le(3x_4)^5\\ 1\le 3x_4 \end{array}$$
Claim 7: If all $x_i$ are positive, $x_1+x_2+x_3+x_4+x_5\le\frac53$. Let $$ u_1=x_1+x_2+x_3\\ u_2=x_2+x_3+x_4\\ u_3=x_3+x_4+x_5\\ u_4=x_4+x_5+x_1\\ u_5=x_5+x_1+x_2\\ s=x_1+x_2+x_3+x_4+x_5\\ $$
By the Generalized Mean Inequality
$$\frac{\Sigma u_i}5 \le (\frac{\Sigma u^5_i}5)^\frac1 5$$
The left hand side is $\frac{3s}5$, and the right hand side is $(\frac{\Sigma 3x_i}5)^\frac15 = (\frac{3s}5)^\frac15$. And $\frac{3s}5 \le (\frac{3s}5)^\frac15$ implies $\frac{3s}5 \le 1$, or $s\le\frac53$.
Now with these seven claims, we can complete our argument.
Let us suppose at least three of $x_1,x_2,x_3,x_4,x_5$ are non-negative: otherwise, replacing each by its negation will also satisfy the equations, with three non-negative. Then by Claim 3, all $x_i$ are non-negative. If one is zero, then by Claim 4, they are all zero. This is one solution. So assume now that all $x_i$ are positive.
Consider the number of $x_i$ in the interval $(0, \frac13]$. If there are at least three, then by Claim 2 all five are in this interval. So by Claim 6, the maximum is $\frac13$. By Claim 5, they are all $\frac13$. This satisfies the original equations, and so this is a second solution.
So now we can assume at most two of the $x_i$ are in $(0, \frac13]$. So there are at least three in $(\frac13, \infty)$. By Claim 1, any remaining $x_i$ are $\ge\frac13$. So $\Sigma x_i\gt\frac53$, contradicting Claim 7.
For a third solution, multiply the previous solutions through by $-1$. So there are only three real solutions: $x_1=x_2=x_3=x_4=x_5=1/3, -1/3, \text{or } 0$.