A homogeneous cubic polynomial with no non-trivial zero over $\mathbf{F_7}$

$$ x^3 + y^3 + z^3 + xyz + xy^2 + 4yz^2 + 2 z x^2 $$

indeed, of the 343 triples $(x,y,z),$ only when we have $(0,0,0)$ is the polynomial equal to $0 \pmod 7.$ Each of the other six values mod 7 is represented 57 times, and $6 \cdot 57 = 342$

Note: the answer recommended by user760870 is something I asked about years ago, at primes represented integrally by a homogeneous cubic form

Alright; this is what they wanted us to figure out. Take a polynomial in one variable that is irreducible over the field, I choose $$ x^3 - 3 x -1 $$

Next, make the 3 by 3 matrix that is the companion matrix for the polynomial.

$$ M = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 3 & 0 \\ \end{array} \right) $$

Finally, take the matrix $$ xI + yM + z M^2 $$

to get the norm form

$$ \det (xI +yM +z M^2)= \color{red}{ x^3 + y^3 + z^3 -3xyz -3yz^2 +6z x^2 -3xy^2 + 9 z^2 x } $$

COMPUTER CAUTION: if using integers on a computer, it is not enough to command t % 7, since the t resulting from the cubic above may be negative, ans t % 7 may be a number from $-6$ to $0.$ My variable was named sump, I was getting wrong totals until I put in

 sump %=7;

sump += 7;

sump %=7;

...........................................

? factormod( x^3 - 3*x-1,7)
%35 = 
[Mod(1, 7)*x^3 + Mod(4, 7)*x + Mod(6, 7) 1]


    ? 
    ? mat = [ 0,1,0; 0,0,1; 1,3,0]
    %31 = 
    [0 1 0]

    [0 0 1]

    [1 3 0]

    ? charpoly(mat)
    %32 = x^3 - 3*x - 1
    ? d = x* matid(3) + y * mat + z * mat^2
    %33 = 
    [x       y       z]

    [z x + 3*z       y]

    [y 3*y + z x + 3*z]

    ? normform2 = matdet(d)
    %34 = x^3 + 6*z*x^2 + (-3*y^2 - 3*z*y + 9*z^2)*x + (y^3 - 3*z^2*y + z^3)
    ? 

..........