That's correct.

The problem I had stumbled on was actually
$$
(a-x)(b-x)(c-x)...(z-x)
$$
but I thought that one would have been too obvious, so I decided to rewrite it in the way I did.

The square roots are all zero because $$(a+2)^2 = (a^2+4a+4)$$.

So are the integrals, because $$x^3$$ is an odd function and the intervals are symmetric around 0.

The logs simplify to $$ln(e^a) = a$$ etc.

Hence, we get

$$
(a-x)(b-x)(c-x)...(x-x)(y-x)(z-x)
$$

Since $$(x-x) = 0$$, the end result must be

$$
0
$$

Looks more complicated than it is (or I'm wrong) :-)

Scroogey

Two options:
- Click the dots and then click on `quote reply`. This should give you the mathjax syntax. You can use that in any program that recognizes `tex` syntex.
- Take a screenshot and ask ChatGPT to give you the mathjax. It's pretty good at doing this, with occasional errors though.

Sorry but an unrelated question I'm asking here. 

**Why can't I copy the equation?**



Coinsreporter

I don't want to spoil it, but I'm pretty sure I have it figured out. 

Aardvark

science

Simplify this expression:

$$
\begin{split}
\left(\sqrt{\frac{(a^2 + 4a + 4) - (a + 2)^2}{\cos^2(x) + \sin^2(x)}} + \ln(e^a) - \int_{-a}^{a} x^3 \, dx - x \right) \times \\ \left(\sqrt{\frac{(b^2 + 4b + 4) - (b + 2)^2}{\cos^2(x) + \sin^2(x)}} + \ln(e^b) - \int_{-b}^{b} x^3 \, dx - x\right) \times \\ \left(\sqrt{\frac{(c^2 + 4c + 4) - (c + 2)^2}{\cos^2(x) + \sin^2(x)}} + \ln(e^c) - \int_{-c}^{c} x^3 \, dx - x\right) \times \\ ... \times \\ \left(\sqrt{\frac{(z^2 + 4z + 4) - (z + 2)^2}{\cos^2(x) + \sin^2(x)}} + \ln(e^z) - \int_{-z}^{z} x^3 \, dx - x\right)
\end{split}
$$

(there are 26 factors in this product)


Previous iteration: https://stacker.news/items/792583/r/south_korea_ln (nearly unsolvable, by design)