How do I prove [a+,[a+,a]]=0 for quantum oscillator operators?

[meanyack]

Homework Statement

Actually the original question is too long but I need help for this tiny part that confuses me. I'll be done if I can show that
[a⁺,[a⁺,a]]=0 and similarly
[a,[a⁺,a]]=0
[a,[a,a⁺]]=0 etc
where a⁺ is the raising and a is the lowering ladder operator in quantum oscillator.

Homework Equations

I tried the formulas
[A,[B,C]]= -[C,[A,B]] -[B,[C,A]] and
a$$\psi$$_n=$$\sqrt{n}$$$$\psi$$_n-1
a⁺$$\psi$$_n=$$\sqrt{n+1}$$$$\psi$$_n+1

The Attempt at a Solution

When I use the formula I found
[a⁺,[a⁺, a]]= -[a,[a⁺,a⁺]]- [a⁺,[a,a⁺]]
On the rhs, first term is zero but I have no idea about the second

Also I tried to use $$\psi$$_n for this operator but it gives something like
[a,[a,a⁺]]$$\psi$$_n=(n^3/2+1)$$\psi$$_n-(n^3/2+n^1/2)$$\psi$$_n-1
which does not seem to be zero, trivially.

 

[Dick]

You are making this way too complicated. Isn't [a,a^(+)]=1? Isn't that what a raising/lowering operator means?