Originally Posted by
ShadowKnight8702 Im not quite sure how that is solved. But one piece of info i forgot to mention is that x is a 12 digits number.
The way to solve it, is to try each modulo class of x.
That's quite doable since there are only 10.
For instance for $\displaystyle x \equiv 3 \pmod{10}$, you get:
$\displaystyle 78^3 \equiv 3 \pmod{10}$
$\displaystyle (78 \pmod{10})^3 \equiv 3 \pmod{10}$
$\displaystyle 8^3 \equiv 3 \pmod{10}$
$\displaystyle 2 \equiv 3 \pmod{10}$
This is a contradiction, so any number x that ends in 3 will not satisfy the equation.