Why is the power of a filament lamp directly proportional to the cube of its voltage? I was doing a textbook question on how the power of a bulb varies with the potential difference across it. I plotted this graph:

(V is on the x axis and P is on the y axis.)
I was then told that this graph obeys the relationship $P=kV^3$ (note that for ohmic conductors the relationship is $P=kV^2$, where $k=\frac1 R$), and I was then told to explain this relationship. I'm not terribly sure where to start; I know that increasing the voltage increases the temperature of the filament and therefore the resistance, but doesn't that mean the current decreases, therefore decreasing the power of the bulb?
For those of you in the British education system, I'm just starting my A level in Physics (which means I'm 17 years old for everyone else), so a decently non-technical answer would be appreciated...
 A: Think of it the following way. You were right in writing the relationship between power, voltage and resistance:
$$P=\frac{V^2}{R}.$$
But this equation was said to not fit the data, and instead
$$P=kV^3.$$
Comparing these two equations, we arrive at the relationship
$$R=\frac{1}{kV}.$$
This is the thing we need to explain. As you've hinted, it's related to the temperature dependence of the material.
Consider the filament to be a black body of area $A$ and at temperature $T$. The electrical power going through it should be irradiated out through black body radiation. We then say
$$P=\sigma AT^4,$$
Where $\sigma$ is the Stefan-Boltzmann constant. Comparing this to the expression given for the power
$$kV^3=\sigma AT^4$$
$$V=\left(\frac{\sigma A}{k}\right)^{1/3}T^{4/3}.$$
Substituting this into our expression for the resistance we get
$$R=\left(\frac{1}{k^2\sigma A}\right)^{1/3} T^{-4/3}.$$
This is quite odd, to be honest. It implies the resistance decreases with temperature, which is quite the opposite of what we have with conductors usually. Are you sure the power goes as $V^3$?
A: A better way to look at this problem is to see that the power used by the tungsten filament is actually proportional to the voltage to the 1.54 power.
The graph you show, has the power proportional to the voltage to the 3.2 power. This is unphysical and typical of physics textbook writers who post hypothetical problems.
The source for this equation and other lighting questions can be found in The "Lighting Handbook: Reference and Application" (8th ed.) from the IES (Illuminating Engineering Society) p.186
