Problem 1601

1. Let $a = \sqrt[3]{pq^2}$, $b = \sqrt[3]{qr^2}$, and $c = \sqrt[3]{rp^2}$. Given that $a + b + c$ is a nonzero rational number, we need to prove that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is also a rational number.

2. Since $a = \sqrt[3]{pq^2}$, $b = \sqrt[3]{qr^2}$, and $c = \sqrt[3]{rp^2}$, we have:
$$a^3 = pq^2, \quad b^3 = qr^2, \quad c^3 = rp^2$$
Therefore, $abc = \sqrt[3]{pq^2} \cdot \sqrt[3]{qr^2} \cdot \sqrt[3]{rp^2} = \sqrt[3]{(pq^2)(qr^2)(rp^2)} = \sqrt[3]{p^3q^3r^3} = pqr$.

3. Given that $a + b + c$ is rational, we also know that $a^3 + b^3 + c^3$ is rational because $a^3 = pq^2$, $b^3 = qr^2$, and $c^3 = rp^2$ are all rational numbers.

4. Using the identity for the sum of cubes:
$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$
Since $a^3 + b^3 + c^3$ and $3abc$ are rational, and $a + b + c$ is rational, it follows that:
$$a^2 + b^2 + c^2 - ab - bc - ca$$
must also be rational.

5. Next, consider the square of the sum $a + b + c$:
$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$
Since $(a + b + c)^2$ is rational and $a^2 + b^2 + c^2 - ab - bc - ca$ is rational, we can solve for $ab + bc + ca$:
$$(a + b + c)^2 - 3(ab + bc + ca) = a^2 + b^2 + c^2 - ab - bc - ca$$
Therefore, $ab + bc + ca$ must be rational.

6. Finally, we need to show that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is rational. Using the identity:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + bc + ca}{abc}$$
Since $ab + bc + ca$ is rational and $abc = pqr$ is rational, it follows that:
$$\frac{ab + bc + ca}{abc}$$
is rational.

Conclusion:
$$\boxed{\frac{1}{\sqrt[3]{pq^2}} + \frac{1}{\sqrt[3]{qr^2}} + \frac{1}{\sqrt[3]{rp^2}} \text{ is a rational number.}}$$