Gaussian curvature of a surface does not take the constant value 1? I came across this very complex equation (calculating the Gaussian curvature of a surface):
\begin{align*}
1 \not\equiv &-\frac{m}{2}\Bigl(\frac{3}{2}C+Su^{-1}-Tu^{-1}+2+Qu^{-1}\Bigr)\\
&\qquad\times(u^3C+Su^2-Tu^2)^{(-2u-2Q-5S+5T)/(6uC+4S-4T)} \\
&+\frac{m}{2}\Bigl[u^3(4C^2-3C)+u^2\Bigl(\frac{1}{3}TC-\frac{1}{3}SC-2S+2T-3QC\Bigr)\\
&\qquad\qquad+u(4TS-2S^2-2T^2-2QS+2QT)\Bigr]\\
&\qquad\times(u^3C+Su^2-Tu^2)^{(-2u-2Q-9S+9T-6uC)/(6uC+4S-4T)}
\end{align*}
I should somehow be able to prove that the right side of the equation is not identically equal to $1$.
I would need some rule or something that "by eye" makes it obvious that it can not be $1$ (i.e. is not a constant function equal to $1$).
$T$, $S$, $Q$, $C$ and $m$ are all constants (all the constants can not be zero, $m>2$ and must be integer, $T$ can not be equal to $S$ ); the only variable is $u$.
I thought that to be worth 1 certainly the functions that are "exponents" must vanish, so they should become zero, but the numerators of the two "exponents" being different, if one is zero (for some value of the constants) the other can not zero and therefore the formula  will never be constant equal to 1.
Tips?
 A: First of all, we introduce $R = T-S$, since $T$ and $S$ only appear in that combination.  Secondly, we introduce the notation
$$
\alpha = \frac{-2 u - 2Q + 5 R}{6 C u - 4 R}
$$
which is the first exponent: the second exponent is $\alpha -1$.
If I have not made an error, I find that the RHS to the expression in the post simplifies to the following:
$$
\frac{m}{2} (C u - R)^{\alpha-1} u^{2\alpha -1} \left( \tfrac52 C(C-2) u^2 + (\tfrac{17}{6} CR + 4 R - 4 CQ) u + 3 R(Q-R)\right),
$$
which we can rewrite in a more suggestive form as follows:
$$
\frac{m}{2} (C u^3 - Ru^2)^{\alpha-1} \left( \tfrac52 C(C-2) u^3 + (\tfrac{17}{6} CR + 4 R - 4 CQ) u^2 + 3 R(Q-R)u\right),
$$
which has the form
$$
\frac{m}{2} P_1(u)^{\alpha -1} P_2(u)
$$
where $P_1$ and $P_2$ are two cubic polynomials in $u$.  Your conditions say that neither $C$ nor $R$ can be zero, so that $P_1(u)$ is not zero for all $u$.  Hence for this expression to be equal to $1$ for all $u$, it must be the case that
$$
P_1(u)^{1-\alpha} = \frac{m}{2} P_2(u)
$$
Since $C \neq 0$, there are two possibilities: either $\alpha = 0$ and then $P_1/P_2 = m/2$ for all $u$, or else $\alpha = 1$ and then we must ensure that $m/2 P_2(u)=1$ for all $u$.  It follows from the expression for $\alpha$ above that it cannot be equal to zero (unless perhaps you are allowed to take the limit $C \to \infty$), so the only possibility is $\alpha = 1$.
This condition fixes $C = -1/3$ and $Q = 9R/2$, and plugging these back into the expression for $P_2$ we see that it is not constant.
I don't discard having made calculational errors, but I think the idea is sound.
