if $x,y,z$be postive real numbers, solve systems of this following equation $$ 3x^3(x+1)^2=2y^2(z+3)^3\tag{1}$$ $$3y^3(y+2)^2=2z^2(x+1)^3\tag{2}$$ $$3z^3(z+3)^2=2x^2(y+2)^3\tag{3}$$
My approach is as follows: $(1)\times(2)\times (3)$,we have $$27xyz=8(x+1)(y+2)(z+3)$$but I have no idea how to continue. Can someone help me?Thanks
Applying the Buchberger algorithm for it over the complex numbers, and assuming the additional equation $$27xyz=8(x+1)(y+2)(z+3),$$ I obtain the following real solutions: Either one of $x,y,z$ is zero, or $$ (x,y,z)=\left(\frac{86}{19},\frac{70}{19},\frac{72}{19}\right). $$ It would be interesting to know the context in which these equations arise, and to ask for all rational solutions of this system of polynomial equations.