How to prove this equals something else. I'm trying to prove that
$$\dfrac{l}{c+u}+\dfrac{l}{c-u}=\dfrac{2l}{\sqrt{c^2-u^2}}$$
My workings so far:
$$\dfrac{l}{c+u}+\dfrac{l}{c-u},$$
put over common denominator,
$$\dfrac{l(c-u)}{(c-u)(c+u)}+\dfrac{l(c+u)}{(c-u)(c+u)},$$
$$\dfrac{l(c-u)+l(c+u)}{(c-u)(c+u)},$$
expand out,
$$\dfrac{cl-lu+cl+lu}{(c-u)(c+u)},$$
group up and cancel out,
$$\dfrac{2cl}{c^2-u^2},$$
we need to make it equal to
$$\dfrac{2l}{\sqrt{c^2-u^2}}$$
Can it actually be done?
 A: The falsity of the equation in the question was historically important for the invention of special relativity theory.  The left side represents the classical (non-relativistic) computation of the time needed for light to travel a distance $l$ to a mirror and back, if the  whole system is moving with speed $u$ in the direction of the light's travel (so the light has effective speed $c+u$ on one leg of the trip and $c-u$ on the other).  The right side is the classical time needed for the same thing if the whole system is moving with velocity $u$ perpendicular to the direction of the light's travel. The fact that these are different motivated the Michelson-Morley experiment, attempting to measure the earth's velocity through the "luminiferous ether", by comparing (via interference effects) the time needed for light to make this back-and-forth trip in different directions.  The result of the experiment, namely that there was no difference in the light's travel times, showed that light does not obey the classical rules for computing its speed relative to moving systems. The ultimate outcome was Einstein's principle that the speed of light is the same, as measured by any inertial observers, regardless of their motion relative to each other, and this is part of the basis of special relativity theory.
