<s>[INST] For positive operators $A$ and $B$ with $A^6=B^6$ show that $A=B$ Since $A$ and $B$ are positive, I managed to show that $A^6$ and $B^6$ are positive.
Now, I can use the fact that there exists a unique square root of both of those and since they're equal, their roots must be equal, so $A^3=B^3$. But what now?
I'm guessing I need to show that $A^2=B^2$ somehow since then I can say 
$A^3=B^3\Rightarrow AA^2=BB^2 \Rightarrow A=B$
How can I do that?
 [/INST] Hint: Subtracting $A^2B$ on both sides leads to $A^2(A-B) = (B^2-A^2)B$. When subtracting $B^2A$ on both sides one obtains $(A^2-B^2)A = B^2(B-A)$. Since $A,B$ are arbitrary, $B-A$ vanishes if and only if $A^2-B^2 = 0$. Then apply the roots and you are done.
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