Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression I'm thinking we could do a contradiction, maybe showing that one of the primes is a composite number if they are in a sequence, but I'm not sure how to finish this up. 
I had this as a math problem in a number theory class, but the deadline has passed so I think it is okay ask.
Suppose not. (We take the negation of the theorem and suppose it to be true.) Suppose there are three cube roots who are in arithmetic progression although not necessarily consecutively.
Let our radicands be the primes $p, q,$ and $r$ and the difference between terms in the arithmetic progression be $d$. $k_1$ and $k_2$ respectively represent the number of common differences. WLOG $p&lt;q&lt;r$.
We can deduce the following statements:
$$\sqrt[3]{r}-\sqrt[3]{q}=k_1d$$
$$\sqrt[3]{q}-\sqrt[3]{p}=k_2d$$
Dividing the first and second equations gives us $k_1\sqrt[3]{p}+k_2\sqrt[3]{r}=(k_1+k_2)\sqrt[3]{q}$. Cubing this equation gives us at least one term with a cube root of $pq$ by the binomial theorem on the LHS, but there are no cube roots on the RHS. This is impossible to satisfy becaues all variables are positive and integral. Because this equation can never be true for primes $p, q$ and $r$, the supposition is false and the statement is true, and we are done.
Suppose not. (We take the negation of the theorem and suppose it to be true.) Suppose there are three cube roots who are in arithmetic progression although not necessarily consecutively.
Let our radicands be the primes $p, q,$ and $r$ and the difference between terms in the arithmetic progression be $d$. $k_1$ and $k_2$ respectively represent the number of common differences. WLOG $p&lt;q&lt;r$.
We can deduce the following statements:
$$\sqrt[3]{r}-\sqrt[3]{q}=k_1d$$
$$\sqrt[3]{q}-\sqrt[3]{p}=k_2d$$
Dividing the first and second equations gives us $k_1\sqrt[3]{p}+k_2\sqrt[3]{r}=(k_1+k_2)\sqrt[3]{q}$. Cubing this equation gives us at least one term with a cube root of $pq$ by the binomial theorem on the LHS, but there are no cube roots on the RHS. This is impossible to satisfy becaues all variables are positive and integral. Because this equation can never be true for primes $p, q$ and $r$, the supposition is false and the statement is true, and we are done.