elementary number theory - If a prime $pmid b$ and $a^2=b^3$, then $p^3mid a$

I have an exercise that I don't know how to solve. I tried to solve it in many ways,
but I didn't get any progress in proving or disproving this...

The exercise is:

Prove or disprove: if $p$ is a prime number, if $a$ and $b$ are native numbers and

$$a^2 = b^3$$ and if $p \mid b$, then
$$p^3 \mid a .$$

If someone has a proof to this exercise I would really appreciate it.

Thanks!

Answer

Let the highest powers of $p$ in $a,b$ be $A(\ge0),B(\ge1)$ respectively,

So, we have $2A=3B\implies \dfrac{2A}3=B\implies 3|2A\implies 3|A\implies A\ge3$