Equations and Functions #1

Let $f(p)=\dfrac{p^3+16}{p}$ .

Then let $k=f(x)=f(y)\neq f(z)$ .

Now you can see it means that the equation $f(p)=k$ has only two solutions, $p=x$ and $p=y$ .

Which also means that graphs $q=f(p)$ and $q=k$ meet at only two points, on an orthogonal coordinates system with a $p$ -axis and a $q$ -axis.

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We know that saying $x< y$ does not harm the generality.

Then $f'(p)=2p-\dfrac{16}{p^2}$ . By simple calculations we can figure out that the point where $f'(p)=0$ is $(2,12)$ .

Now let's draw the graph. By differentiating once and twice, we get the graph like below.

Seeing the graph above, we now can figure out that $q=k$ passes through $(2,12)$ .

Therefore $k=12$ .

Then $x$ and $y$ are two distinct roots of $f(p)=\dfrac{p^3+16}{p}=12$ .

$p^3-12p+16=0$

$(p-2)^2(p+4)=0$

Therefore, $x=-4$ and $y=2$ , leading to $\boxed{(x-y)^2=36}$ .