I encountered the following problem.
Let $$P(x) = 2x^3 − 2ax^2 + (a^2 − 81)x − b$$ For $a, b \in \mathbb{N}$, $P(x) = 0$ has $3$ positive integer roots. The value of $a$ is unique, but it yields two distinct values for $b$. Find both values of $b$.
What I did was first apply Vieta's formulas to obtain
$$\begin{align*} a &= p + q + r \\ \frac{a^2-81}{2} &= pq + pr + qr \\ \frac{b}{2} &= pqr \end{align*}$$ where $p,q,r$ are the roots.
The second equation yields $$a^2 = 2(pq + pr + qr) + 81 \quad (1)$$ and squaring the first produces $$a^2 = p^2+q^2+r^2+2(pq + pr + qr) \quad (2)$$
Equating $(1)$ and $(2)$, we know that $$p^2+q^2+r^2=81$$
Because $p,q,r$ are positive integers, we can calculate them by brute force. What I got was, without loss of generality, $$p=1, q=4, r=8$$$$p=3,q=6,r=6$$$$p=4,q=4,r=7$$
Using the last equation from the original set of Vieta's formulas produces $b=64,216,$ or $224$. However, the answer is supposedly only $216,224$. Why is $64$ not a solution? Did I do something wrong?
Yes you have done everything correctly but, it is given that value of $a$ is unique. So if you take all cases into account, value of a won't be unique. $Case \ 1: a=13$ $Case \ 2:a =15$ $Case\ 3:a=15$ So to get $2$ values of b you should choose $a=15$. Therefore consider only the last 2 solutions.