Find the maximum of the value $F=x^3y+y^3z+z^3x$ let $x,y,z$ be real number.if $x+y+z=3$,show that
$$x^3y+y^3z+z^3x\le \dfrac{9(63+5\sqrt{105})}{32}$$
and  the inequality $=$,then $x=?,y=?,z=?$
I can solve if add $x,y,z\ge 0$,also see:  Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y, z \ge 0$.
But for real $x,y,z$ I can't solve it


It is not surprizing that Mathematica does not produce explicit expressions for $(x,y,z)$.
Using it with the restriction $x>y>z$, it gives that $y$ and $z$ are respectively  the fourth and the first roots of the equation
$$56 t^6-336 t^5+378 t^4+819 t^3-2079 t^2+1512 t-351=0$$
Using  @River Li's comment, $(x,y,z)$ are the solutions of the cubic
$$u^3-3u^2-\frac{3}{8} \left(3+\sqrt{105}\right)u+\frac{3}{112} \left(147+17 \sqrt{105}\right)=0$$ and, using the trigonometric method, they are
$$x=1+\sqrt{\frac{11+\sqrt{105}}{2} } \cos \left(\frac{1}{3} \cos
   ^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2}
   \left(11+\sqrt{105}\right)^{3/2}}\right)\right)=3.71382\cdots$$
$$y=1-\sqrt{\frac{11+\sqrt{105}}{2} } \sin \left(\frac{\pi
   }{6}-\frac{1}{3} \cos ^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2}
   \left(11+\sqrt{105}\right)^{3/2}}\right)\right)=1.20642\cdots$$
$$z=1-\sqrt{\frac{11+\sqrt{105}}{2} } \sin \left(\frac{\pi
   }{6}+\frac{1}{3} \cos ^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2}
   \left(11+\sqrt{105}\right)^{3/2}}\right)\right)=-1.92024\cdots$$
Trying to convert to radicals would be more than problematic.
Edit
After a series of mistakes, using @River Li's comments, the factorization of the sextic as the product of two cubic polynomials
$$(t^3 + b_2t^2 + b_1t + b_0)\,(t^3 + a_2t^2 + a_1t + a_0)$$gives
$$b_2=-3 \qquad \qquad b_1=-\frac{3}{8} \left(3-\sqrt{105}\right)\qquad \qquad b_0=\frac{3}{112} \left(147-17 \sqrt{105}\right)$$
$$a_2=-3 \qquad \qquad a_1=-\frac{3}{8} \left(3+\sqrt{105}\right)\qquad \qquad a_0=\frac{3}{112} \left(147+17 \sqrt{105}\right)$$
