Does $x^3 \equiv 6 \pmod {11}$ have any solutions?

Question

My gut instinct is that it does not, but I am unsure of how to show this... and I think the $x^3$ is what is causing me trouble in figuring this out. I have attempted to rewrite it using the definition of divisibility and proceeding to $x^3-6=11k$, where $k$ is an integer. However, at this point I am unsure of how to proceed. What would the next step be, or is this the wrong way to approach this problem?

Answer 1

The solution is $$x=8 \mod{11}$$ We know it is unique by finding the values of $x^3 \mod{11}$ for $0 \le x \lt 11$ as we can reduce all other values of $x$ I to this range.

Answer 2

Make a table of cubes. As it is obvious that $0,1,-1$ are not solutions, we'll make a table for the other values. Note that $6\equiv -5\mod 11$: \begin{array}{l|rrrr|} x&\pm2&\pm3&\pm 4&\pm 5 \cr \hline x^2&4&-2&5&3\cr x^3& \mp3&\pm 5&\mp2&\pm4 \cr \hline \end{array}

so the solutions are $\;x\equiv -3\;\text{ (or }8)\mod 11$.

Answer 3

$\bmod 11\!:\ \left[x^{\large 3}\equiv 6\right]^{\large -3}\!\!\!\!\!\!\Longrightarrow\, \underbrace{x^{\large -9}}_{\Large x}\equiv \left(\dfrac{1}6\right)^{\large 3}\!\!\equiv \left(\dfrac{12}6\right)^{\large 3}\!\!\equiv 8$