Somewhat basic diophantine equation

We see that the trivial solution is $(a,b,c)=(0,0,0)$ . If two of $a,b,c$ are zero, it implies that the another one must be zero as well. If one of $a,b,c$ is zero, there will be no integral solution but all zero, since $\sqrt[3]{3}, \sqrt[3]{5}, \sqrt[3]{\frac{5}{3}}$ are not rational. Therefore, it is left to find solution that none of $a,b,c$ is zero.

Because this equation is homogeneous (of order $3$ ), we assume that $\gcd(a,b,c)=1$ . Notice that $5 \mid a$ , so we let $a=5x$ , and get the new (reduced) equation $75x^3+b^3+3c^3 = 45xbc.$ Again, we see that $3 \mid b$ , so we let $b=3y$ , and have the new equation $25x^3+9y^3+c^3 = 45xyc.$ Take modulo $9$ , we have that $7x^3+c^3 \equiv 0 \pmod{9}.$ Since the possible residues of a cube modulo $9$ are $0,1,8$ , we write the table of $7x^3+c^3 \mod{9}$ as follow. $\hspace{70pt} (x^3 \mod{9})\\ \hspace{75pt} ~~ ~~ 0 ~~ ~~ ~~ 1 ~~ ~~ ~~ 8 ~~ ~~\\ \hspace{60pt} ~~ 0 ~~ || ~~ 0 ~~ || ~~ 7 ~~ || ~~ 2 ~~ ||\\ (c^3 \mod{9}) ~~ ~~ 1 ~~ || ~~ 1 ~~ || ~~ 8 ~~ || ~~ 3 ~~||\\ \hspace{60pt} ~~ 8 ~~ || ~~ 8 ~~ || ~~ 6 ~~ || ~~ 1 ~~||$ We see that the only case that makes $7x^3+c^3 \equiv 0 \pmod{9}$ is that $x^3 \equiv 0 \pmod{9}$ and $c^3 \equiv 0 \pmod{9}$ . Finally, we get that $3 \mid x$ and $3 \mid c$ . This means that $a,b,c$ are all divisible by $3$ , contradicting our assumption that they are relatively prime. Thus, there is no other solution than $(a,b,c)=(0,0,0)$ , and the answer to this problem is $0+0+0=\boxed{0}$ .