$4$ variable system of equations Find all the solutions to this system of equations:
$$\begin{cases} 
-2d^3+3a^2d+d+2c^3-3a^2c-c=0\\
2d^3-3db^2-d-2c^3+3b^2c+c=0\\
-6c^2b+b^3+b+6c^2a-a^3-a=0\\
6d^2b-b^3-b-6d^2a+a^3+a=0
\end{cases}$$
After factoring:
$$\begin{cases} 
d(-2d^2+3a^2+1)+c(2c^2-3a^2-1)=0\\
d(2d^2-3b^2-1)-c(2c^2-3b^2-1)=0\\
b(-6c^2+b^2+1)+a(6c^2-a^2-1)=0\\
b(6d^2-b^2-1)-a(6d^2-a^2-1)=0
\end{cases}$$
How should I proceed next?
System of equation comes from following problem:
Find the line where line integral
$$\int_L (y^3-y)dx+2x^3dy$$
is the largest
From Green's theorem:
$$\begin{cases} F_x=6x^2 \\ Q_y=3y^2-1 \end{cases}$$
$$\underset{D\ \ }{\iint}(6x^2-3y^2-1)dxdy = \int_{a}^{b}(2x^3-3y^2x-x)\Bigg|_{c}^{d}dy=\int_{a}^{b}((2d^3-3y^2d-d)-(2c^3-3y^2c-c))dy=$$
$$=\left((2d^3y-y^3d-dy)-(2c^3y-y^3c-cy)\right)\Bigg|_{a}^{b}=$$
$$=\left((2d^3b-b^3d-db)-(2c^3b-b^3c-cb)\right)-\left((2d^3a-a^3d-ad)-(2c^3a-a^3c-ac)\right)=$$
$$=2d^3b-db^3-db-2c^3b+b^3c+cb-2d^3a+a^3d+ad+2c^3a-a^3c-ac$$
I need to find $t(a,b,c,d)$ partial derivatives:
$$t_a=-2d^3+3a^2d+d+2c^3-3a^2c-c$$
$$t_b=2d^3-3db^2-d-2c^3+3b^2c+c$$
$$t_c=-6c^2b+b^3+b+6c^2a-a^3-a$$
$$t_d=6d^2b-b^3-b-6d^2a+a^3+a$$
Now I equal all of them $0$, and find global maximum.

Adding $1$st and $2$nd equations, one can get
$$
3d(a^2-b^2)-3c(a^2-b^2)=0,
$$
$$
(d-c)(a-b)(a+b)=0\tag{1};
$$
adding $3$rd and $4$th equation, one can get
$$
b(6d^2-6c^2)-a(6d^2-6c^2)=0,
$$
$$
(b-a)(d-c)(d+c)=0\tag{2}.
$$
Equations $(1), (2)$ can help here?
Other (similar) way:
write system in form: 
$$\begin{cases} 
2(d^3-c^3)-3a^2(d-c)-(d-c)=0\\
2(d^3-c^3)-3b^2(d-c)-(d-c)=0\\
6c^2(b-a)-(b^3-a^3)-(b-a)=0\\
6d^2(b-a)-(b^3-a^3)-(b-a)=0\\
\end{cases}$$
$$\begin{cases} 
(d-c)(2d^2+2dc+2c^2-3a^2-1)=0\\
(d-c)(2d^2+2dc+2c^2-3b^2-1)=0\\
(b-a)(6c^2-b^2-ab-a^2-1)=0\\
(b-a)(6d^2-b^2-ab-a^2-1)=0\\
\end{cases}$$
$$\begin{cases} 
(d-c)(2d^2+2dc+2c^2-3a^2-1)=0\\
(d-c)(b-a)(b+a)=0\\
(b-a)(6c^2-b^2-ab-a^2-1)=0\\
(b-a)(d-c)(d+c)=0\\
\end{cases}$$
Cases:


*

*$a=b$, $c=d$: then system has solution for any $a,c$;

*$a=b$, $c\ne d$: then remains $1$ equation: $2(d^2+dc+c^2)-3a^2-1=0$;

*$a\ne b$, $c=d$: then remains $1$ equation: $6c^2-(a^2+ab+b^2)-1=0$;

*$a\ne b$, $c\ne d$: then $b=-a$, $d=-c$, and 
$$
\begin{cases}
2c^2-3a^2=1;\\
6c^2-a^2=1;\\
\end{cases}
$$
$\qquad$(linear system on $a^2, c^2$).  
