Equations and Functions #2

Multiply $b^2e^a$ to the original expression and we get:

$c^2e^c\neq b^2e^b=a^2e^a$ .

Then we see that it does not harm the generality to say that $a<b$ .

Let $f(x)=x^2e^x$ , and $a^2e^a=b^2e^b=k$ .

We see that the question originally meant that $f(x)=k$ has only two solutions.

Drawing the graph $y=f(x)$ shows that $f''(a)=0$ and $a\neq0$ , and the solution to it is $a=-2$ .

Therefore, $b^2e^b=\dfrac{4}{e^2}$ . Multiply $e^2$ to both sides and we get $b^2e^{b+2}=4$ .

$\dfrac{(abe)^4}{e^{ab}}=a^4b^4e^{4-ab}=16b^4e^{2b+4}=16\left(b^2e^{b+2}\right)^2=16\cdot4^2=\boxed{256}$ .

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Note: If you've got any problems on understanding this solution, refer to the problem <Equations and Functions #1> , and read carefully through the solution. The basis of this problem is the same as the linked problem!