1. The problem is to find a 3x3 determinant that results in a cubic polynomial in the variable $x$. 2. Recall the formula for the determinant of a 3x3 matrix: $$\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$$ where the matrix is $$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$$ 3. To get a cubic polynomial in $x$, we can include $x$ terms in the matrix entries such that the determinant expands to a cubic expression. 4. Consider the matrix: $$\begin{bmatrix} x & 1 & 0 \\ 0 & x & 1 \\ 1 & 0 & x \end{bmatrix}$$ 5. Calculate the determinant: $$\det = x(x \cdot x - 1 \cdot 0) - 1(0 \cdot x - 1 \cdot 1) + 0(0 \cdot 0 - x \cdot 1)$$ $$= x(x^2 - 0) - 1(0 - 1) + 0(0 - x)$$ $$= x^3 + 1$$ 6. The determinant is $x^3 + 1$, which is a cubic polynomial in $x$. Final answer: $$\det = x^3 + 1$$