If $a^3 \equiv b^3\pmod{11}$ then $a \equiv b \pmod{11}$. If $a^3 \equiv b^3\pmod{11}$ then $a \equiv b \pmod{11}$.


I verified the above statement for smaller values of $a$ and $b$ but I am unable to prove it in general. Please prove or disprove it. Can we replace $11$ by any other integer?
Here is a general statement:


  If $p$ is a prime and $m$ is coprime with $p-1$, then $a^m \equiv b^m \bmod p$ iff $a \equiv b\bmod p$.


This follows from Fermat's theorem by writing $(p-1)(-u)+mv=1$ for $u,v \in \mathbb N$.

In your example, $p=11$ and $m=3$. We can take $u=2$ and $v=7$ to get $10\cdot (-2) + 3\cdot(7)=1$. Write this as $1 + 10 \cdot 2 = 3\cdot 7 $. Then
$$
a \equiv a \cdot a^{10 \cdot 2} = a^{1 + 10 \cdot 2} = a^{3\cdot7}
\equiv
b^{3\cdot7} = b^{1 + 10 \cdot 2} = b \cdot b^{10 \cdot 2} \equiv b
\bmod 11
$$
Note $$1^{3} \equiv 2^{3}\pmod{7}$$ doesnt imply $1\equiv 2\pmod{7}$