cov(U,V)=E((U-u)(V-v)) where u,v are E(U),E(V) respectfully.
u=E(U)=E(X)=0, for obvious reasons.
So we have:
cov(U,V)=E(U(V-v))
=E(UV-Uv)
=E(UV)-E(Uv)
=E(X^3)-vE(U)
=E(X^3)-vu
=E(X^3)-v.0
=E(X^3)
=0
this last step can be done: x^3 is an odd function, f(x) is an even function, so when we multiply them we get an odd function. Hence the integral of x^3.f(x) (ie the expected value of X^3) is 0.
Dependence follows...
Some helpful hints with this stuff:
Expected values are one of your most useful tools with probability theory:
Know stuff like:
A random variable can never equal an expected value unless the random variable is constant. So if you end up proving something like:
E(X)=Y where X,Y are random variables, you are probably wrong.
Expected value satisfies:
E(aX+bY)=aE(X)+bE(Y)
It DOES NOT SATISFY E(XY)=E(X)E(Y) unless X,Y are independent.
If A is an event, and 1{A} is the indicator function of the event (i.e 1{A} = 1 if event happens, 0 otherwise), then:
E(1{A})=P(A)
This is slightly more advanced, but can often be a useful trick.