Observe the identity $$2(a^5+b^5+c^5)=5abc(a^2+b^2+c^2),$$subject to condition $a+b+c=0$. Proof: It is true under the condition that $a+b+c=0$ that $a^3+b^3+c^3=3abc$. Hence $$(a^3+b^3+c^3)(a^2+b^2+c^2)=3abc(a^2+b^2+c^2).$$We transform the LHS, obtaining $$LHS=a^5+b^5+c^5+a^2b^2(a+b)+b^2c^2(b+c)+c^2a^2(c+a)$$$$=a^5+b^5+c^5-abc(ab+bc+ca)=3abc(a^2+b^2+c^2)$$$$a^5+b^5+c^5=abc(3a^2+3b^2+3c^2+ab+bc+ca)$$But $-2(ab+bc+ca)=a^2+b^2+c^2$, hence the result follows. First, $abc=30$. We take advantage of the identity, giving $a^2+b^2+c^2=38\iff ab+bc+ca=-19$. And we finally see that $a,b,c$ are solutions to $$x^3-19x-30=0$$Giving that $\boxed{(a,b,c)=(-3,-2,5), \text{and its permutations}.}$