<s>[INST] I want to find out the angle for the expression $a^3 + b^3 = c^3$. like in pythagorean theorem angle comes 90 degree for the expression $a^2 + b^2 = c^2$, however I know that no integer solution is possible.
 [/INST] Expanding on Brian M. Scott's answer, since
$\cos\theta=\frac{a^2+b^2-c^2}{2ab}$
and $c^3 = a^3+b^3$,
$\cos\theta=\frac{a^2+b^2-(a^3+b^3)^{2/3}}{2ab}
= \frac{1+(b/a)^2-(1+(b/a)^3)^{2/3}}{2}
=\frac{1+r^2-(1+r^3)^{2/3}}{2}
$
where $r = b/a$.
The derivative of the numerator
is $2r-(2/3)(3r^2)(1+r^3)^{-1/3} = 2r - 2r^2(1+r^3)^{-1/3}$
which is never 0 (else $(1+r^3)^{1/3} = r$)
so is always positive (since its value at 1 is 
$2-2/2^{1/3} > 0$).
For large r, 
$(1+r^3)^{2/3} = r^2 (1+r^{-3})^{2/3}
\approx r^2(1 + (2/3)r^{-3})
= r^2 + 2/(3r)
$
so 
$\cos\theta \approx \frac{1+r^2 - (r^2 + 2/(3r))}{2}
= \frac{1-2/(3r)}{2}
$
which tends to 1/2 for large $r$.
This may have an error, since I would expect it to go to 1,
but I have to go now, so this this is it.
</s>