Positive integer $n$ is divisible by a cube, if it has three prime divisors and at least seven divisors $p^k$ ($p$ prime and $k$ positive integer). Problem. Let $n$ be a positive integer that has exactly three prime divisors, and
at least seven divisors of the form $p^k$, where $p$ is a prime, and $k$ is
a positive integer. Prove that $n$ must be divisible by the cube of an
integer that is larger than 1.
This is a problem (Quick Check 3 Chapter 1.2) from A Walk Through Combinatorics by Miklós Bóna, and comes after the discussions on the Pigeon-hole principle and Generalized Pigeon-hole principle.
(Note: I don't have much background on Number Theory and also unsure if my attempts even lead to the correct solution)

The following are my current attempts and progress, and also some clarifications on the problem:

*

*Let $\alpha$, $\beta$, and $\gamma$ be the three prime divisors of $n$, then
$$n=\alpha \ell, \quad n=\beta \ell, \quad \text{and} \quad n=\gamma \ell$$
for some integers $\ell$ which are not necessarily the same for each instance. Now, I think it follows from the prime factorization theorem that
$$n = (\alpha \beta \gamma)\cdot \ell$$
for some integer (I am unsure if this actually follows).

*Also, $n$ has at least seven prime divisors $p^k$, so for $i\in [7]$
$$n = p_i^{k_i}\cdot \ell_i$$
I tried to equate each of these prime divisors, for example $p_1^{k_1}\cdot \ell_1 = p_2^{k_2}\cdot \ell_2$. Then, since the factor $p_1^{k_1}$ from the left side may not have came from $p_2^{k_2}$ on the right then it must have come from $\ell_2$ (and vice-versa); generalizing this, I think it must follow that
$$n=\left(\prod_1^7  p_i^{k_i} \right)\cdot c $$
for some integer $c$.

*Continuing from the two previous points
$$(\alpha\beta\gamma)\cdot \ell = \left(\prod_1^7  p_i^{k_i} \right)\cdot c$$
From this point, I no longer know what to do. Do we then assume the contrary? That there is no cube which divides $n$? Also, can we assume from the problem that the three prime divisors of $n$ must be distinct from each other, and also distinct from the seven $p^k$ divisors?
 A: Assume the opposite, and arrive at a contradiction.
If $n$ has exactly three prime divisors, it is of the form $n=p_1^{k_1}p_2^{k_2}p_3^{k_3}$. Now assume that no cube divides $n$. That means each $k_i \in \{1,2\}$, lest $p_i^3$ (a cube) is a divisor of $n$.
In this case, the divisors of the form $p^k$ are just $p_1,p_1^2,p_2,p_2^2,p_3,p_3^2$. There are only $6$ of them. But the probem requires that there be at least $7$ divisors of that form, so at least one of the $p_i$ must have an exponent $k_i \ge 3$. Hence, $n$ must be divisible by a cube of that $p_i$, and $p_i$ is a prime number larger than $1$.
A: Let's make this simpler.

*

*$n$ has exactly $3$ prime divisors.

Let them by $p,q,r$ so $n = p^aq^br^c$ for natural numbers $a,b,c$.

*

*$n$ has at least $7$ divisors of the for $p^k$ where $k \ge 1$.

That may seem complex but $p^1, p^2, p^3, ...., p^a$ are exactly $a$ such divisors.  $q^1, q^2, ...., q^b$ are $b$ more and $r^1,...., r^c$ are $c$ more.  And those are all such divisors.  And other divisors of $p^aq^br^c$ must be either $1$ or some combination of at least two primes.
So $n$ has exactly $a+b+c$ such divisors.  So all this is saying is $a+b+c \ge 7$.

*

*$n$ is divisble by a cube larger than $1$.

At this point, let's ask how can that not be possible.
We should note that if $a \ge 3$ then $n=p^aq^br^c = p^3\times p^{a-3}p^bq^c$ so $p^3|n$ and a perfect cube divides $n$.
Likewise if $b\ge 3$ or $c \ge 3$ then $q^3$ or $r^3$ will be a perfect cube factor.
So in order to have no perfect cubes (other than $1$) factors we must have $a < 3; b < 3; c< 3$.
But $a+b + c \ge 7$.
.......
A: I’d state the solution using the pigeonhole principle.  Let $n=p_1^{k_1} p_2^{k_2} p_3^{k_3}.$. Then any number of the form $p_i^k, 1 \leq k \leq k_i$, divides $n$.
There are at least $7$ such divisors and only $3$ primes to choose from, so for some $j$ we must have $p_j^k$ for at least three different positive integers $k$, which means $p_j^k$ divides $n$ for some $k \ge 3$.  But that, in turn, means $p_j^3$ divides $n$ and we are done.
