Assume that $p<q<r$ are primes with $2q^3=p^3+q^3=(p+q)(p^2-pr+r^2)$. Obviously none of these is $=2$. By this $p+q$ is even and we get that one of the following cases happens: a) $p+r=2$: nonsense. b) $p+r=2q$: then $p,q,r$ is an arithmetic progression and $4(p^3+r^3) = (2q)^3 = (p+r)^3 \iff p^3+r^3 = pr(p+r) \iff (p-r)^2 = 0$, impossible. c) $p+r=2q^2$: then $p^2-pr+r^2=q$ and thus $(p+r)^2 > p+r = 2q(p^2-pr+r^2) > 4p^2-4pr+4r^2$. This yields $2pr > p^2+r^2$, nonsense. d) $p+r=2q^3$: then $4=4p^2-4pr+4r^2=(2p-r)^2+3r^2$, wrong by already $3r^2>4$.

Your solution is very nice. Mine is: Suppose it is true. So, we can have $ \sqrt [ 3 ] {p_1} =a$, $ \sqrt [ 3] {p_2} =a +m\cdot d$, $ \sqrt [ 3] {p_3} =a+n \cdot d$, where $ p_1,p_2, p_3$ three different primes, and $ a,d \in N$ . After eliminaton of $ a$ and $ d$, we have $ m \cdot \sqrt [ 3 ] {p_3} - n \cdot \sqrt [ 3 ] {p_2} = (m - n )\cdot \sqrt [ 3 ] {p_1}$, the cube of which is $ m^3\cdot p_3 - n^3 \cdot p_2 - 3 m\cdot n \sqrt [ 3 ] {p_2 \cdot p_3} (m\cdot \sqrt [ 3 ] {p_3} - n \cdot \sqrt [ 3 ] {p_2})= (m - n )^3 \cdot p_1$. Finally, $ \sqrt [ 3 ] {p_1 \cdot p_2 \cdot p_3} = \frac{m^3 \cdot p_1 - n^3 \cdot p_2 - (m - n )^3 \cdot p_1}{ 3 \cdot m \cdot n ( m - n )}$, so contradiction, since $ \sqrt [ 3 ] {p_1 \cdot p_2 \cdot p_3}$ is irrational.