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.

south_korea_ln

science

[Not-so-daily puzzle] Simplify the expression

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) :-)