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Let's $K$ be a ring, and for two elements of the ring we say $$[a,b]=ab-ba.$$

$x,y,h$ are elements of $K$ satisfying: $$[h,x]=2x ,\quad [x,y]=h.$$

Is it possible to prove the following identity: $[h,x^n]=2n.x^n$?

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    Why do you expect this equality to hold? Have you seen it as an exercise? Have you realized it should be true by testing many examples? Or another reason? – 2017-02-16
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    It's just an exercice, and after basic expansion and recursion try out, I'm not sure how to go further to get the proof. – 2017-02-16
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    Can you prove it for $n=2$? – 2017-02-16
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    I try without success. Not sure how to expand further $[h,x^2]=hx^2-x^2h$ and i also try to expand $ [h,x][h,x] =hxhx-hx^2h-xh^2+xhxh$, but i'm still block. – 2017-02-16
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    Let's write $[h,x^2] = hx^2 - x^2h = (hx^2 - xhx) + (xhx - x^2h)$. Can you continue from there? – 2017-02-16
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    I got: $[h,x^2]= [h,x]x+x[h,x]=2x^2+x2x$. But is there a reason to commute $x2x$ into $2x^2$, in which case we get the proof for n=2? – 2017-02-16
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    Yes, if we use the distributivity, $2x= (1+1)x=x+x$ then we ended with $x2x=x(x+x)=x^2+x^2=2x^2$. Thank you for your hints. I will try to get the proof for n. – 2017-02-16
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53800/discussion-between-xaving-and-pierre-guy-plamondon). – 2017-02-17

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Use the rule $$ \mathbf{\color{red}{[a, b c] =}}\ a b c - b c a = a b c - b a c + b a c - b c a\ \mathbf{\color{red}{= [a, b] c + b [a, c]}}, $$ and then induction $$ [h, x^{n}] = [h, x^{n-1} x] = [h, x^{n-1}] x + x^{n-1} [h, x] = 2 (n-1) x^{n-1} x + x^{n-1} \cdot 2 x = 2 n x^{n}. $$