This problem was proposed by me

$a=b=c$ The equations all become $2a^3=a^2\iff 2a^3-a^2=0\iff a^2(2a-1)=0.$ Therefore, $a=b=c=0$ and $a=b=c=\frac12$ work. Two solutions are $\boxed{(0,0,0)}$ and $\boxed{\left(\frac12,\frac12,\frac12\right)}.$ $a=b$ First equation becomes $a^3+a^2c=ac.$ Second equation becomes $a^3+ac^2=a^2.$ Third equation becomes $2a^3=ac.$ Clearly $a=0$ works, and $c$ can be anything. Therefore another solution is $\boxed{(0,0,n)\forall n\in\mathbb R}$ and permutations. Now suppose $a\neq0.$ Then the first equation and the second equation both give $a^2+c=a\iff c=a-a^2.$ The third equation gives $2a^2=c.$ Therefore, we must have $2a^2=a-a^2\iff3a^2=a.$ Since $a\neq0,$ we can divide both sides by $3a$ to get $a=\frac13.$ This value of $a$ gives $c=\frac29.$ Thus, another solution is $\boxed{\left(\frac13,\frac13,\frac29\right)}$ and permutations. Now, $a\neq b\neq c.$ Note that $a,b,c\neq0.$ In the first equation, we have $a^3=ac-b^2c=c\left(a-b^2\right)\iff c=\frac{a^3}{a-b^2}^*.$ Substituting this into the other equations gives $b^3+\left(\frac{a^3}{a-b^2}\right)^2a=ba$ and $\left(\frac{a^3}{a-b^2}\right)^3+a^2b=\left(\frac{a^3}{a-b^2}\right)b.$ In the first equation we get $b^3+\frac{a^7}{\left(a-b^2\right)^2}=ab\iff\frac{b^3\left(a-b^2\right)^2a^7}{\left(a-b^2\right)^2}=ab\iff b^3\left(a-b^2\right)^2+a^7=ab\left(a-b^2\right)^2\iff ab\left(a-b^2\right)^2-b^3\left(a-b^2\right)^2=a^7\iff\left(a-b^2\right)^2=\frac{a^7}{ab-b^3}.$ The second equation gives $\frac{a^9}{\left(a-b^2\right)^2}+a^2b=\frac{a^3b}{a-b^2}\iff\frac{a^9+a^2b\left(a-b^2\right)^2}{\left(a-b^2\right)^2}=\frac{a^3b}{a-b^2}\iff a^7+b\left(a-b^2\right)^2=ab\left(a-b^2\right)^2\iff a^7=ab\left(a-b^2\right)^2-b\left(a-b^2\right)^2\iff\left(a-b^2\right)^2=\frac{a^7}{ab-b}.$ Therefore, we have $\frac{a^7}{ab-b^3}=\frac{a^7}{ab-b}\iff ab-b^3=ab-b\iff b^3=b.$ This means $b=0$ or $b=1.$ However, $b=0$ contradicts the statement $a,b,c\neq0,$ so we can discard this. Thus, we have $b=1.$ Substituting this into $b^3+\left(\frac{a^3}{a-b^2}\right)^2a=ba$ gives $1+\left(\frac{a^3}{a-1}\right)^2a=a\iff1+\frac{a^7}{(a-1)^2}=a\iff\frac{a^7+(a-1)^2}{(a-1)^2}=a\iff a^7+(a-1)^2=a(a-1)^2.$ Expanding gives $a^7+a^2-2a+1=a^3-2a^2+a\iff a^7-a^3+3a^2-3a+1=0.$ *When $a\neq b^2.$ If $a=b^2$ we have $a^3+ac=ac\iff a=0.$ But if we substitute this we get $a=b=c=0,$ contradicting the statement that $a\neq b\neq c$ and also that $a,b,c\neq0.$ I'm stuck here... not sure how to solve that 7th degree polynomial

Here is an easier route: Clearly if one of the variables is $0$, then all of them are $0$ and we get the solution $\boxed{(0,0,0)}$. From now on suppose $a,b,c \ne 0$. From the first equation we get $a-b^2=\frac{a^3}{c}$ while from the second one we get $a-b^2=\frac{c^2a}{b}$ so that we have found $\frac{c}{b}=\frac{a^2}{c^2}=\frac{b^4}{a^4}=\frac{c^8}{b^8}$ (by iterating). Hence $b^7=c^7$ and hence $b=c$ and hence $a=b=c$ and hence the only further solution $\boxed{\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)}$.

My idea for transformation yielding a solution quite similar to Tintarn's is the following: $$a(a^2b-c^3)=b(a^3+b^2c)-c(b^3+c^2a)=b\cdot ac-c\cdot ba=0.$$