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.
 A: And here's a plot of the angle:

The minimum angle seems to be at $a = b$, where
$$\cos\theta = 1 - \frac{1}{2^{1/3}}$$
A: There is no single angle corresponding to the relationship $a^3+b^3=c^3$.
Suppose that a triangle has sides of lengths $a,b$, and $c$ such that $a^3+b^3=c^3$. We know from the law of cosines that if $\theta$ is the angle opposite the side of length $c$, then $c^2=a^2+b^2-2ab\cos\theta$, so
$$\cos\theta=\frac{a^2+b^2-c^2}{2ab}\;.$$
Now let’s look at just a few examples. If $a=b=1$, then $c=\sqrt[3]2$, and $$\cos\theta=\frac{2-2^{2/3}}2\approx0.20630\;.$$
If $a=1$ and $b=2$, then $c=\sqrt[3]9$, and $$\cos\theta=\frac{5-9^{2/3}}4\approx0.16831\;.$$
If $a=1$ and $b=3$, then $c=\sqrt[3]{28}$, and $$\cos\theta=\frac{10-28^{2/3}}6\approx0.12985\;.$$
As you can see, these values of $\cos\theta$ are all different, so the angles themselves are also different.
A: You do get slightly different answers for the Pythagorean Theorem on the surface of a sphere of radius $1,$ or the hyperbolic plane of curvature $-1.$ On the sphere, a right triangle with geodesic lengths of legs $a,b$ and hypotenuse $c$ obeys
$$ \cos c = \cos a \; \cos b,   $$
while in the hyperbolic plane with curvature $-1$ it becomes
$$   \cosh c = \cosh a \; \cosh b.  $$
In both cases, if you write out the functions as power series in $a,b,c,$ you see that the limit as $a,b,c$ all shrink to nearly $0$ is the traditional Pythagorean Theorem. 
See Wikipedia Section Link
A: 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.
