No cube can end with exactly two zeros.

There is no perfect cube which ends in $4$.

For an integer $a, a^3$ is always greater than $a^2$.

If $a$ and $b$ are integers such that $a^2>b^2$, then $a^3>b^3$

If $a$ divides $b$, then $a^3$ divides $b^3$

If $a^2$ ends in $9$, then $a^3$ ends in $7$

If $a^2$ ends in $5$, then $a^3$ ends in $25$

If $a^2$ ends in an even number of zeros, then $a^3$ ends in an odd number of zeros.

Find the cube of:

$-11$