$\frac{d^2}{dx^2} \left\{ \det \begin{bmatrix} x^3 & x \\ 2 & e^x \end{bmatrix} \right\}$ equals

1. Calculate the Determinant

First, we find the determinant of the $2 \times 2$ matrix by using the formula $\det(A) = ad - bc$:

$f(x) = \det \begin{bmatrix} x^3 & x \\ 2 & e^x \end{bmatrix} = (x^3 \cdot e^x) - (x \cdot 2)$

$f(x) = x^3 e^x - 2x$

2. Find the First Derivative

Now, we differentiate $f(x)$ with respect to $x$. For the first term, we apply the Product Rule $(\frac{d}{dx}[uv] = u'v + uv')$:

$f'(x) = \frac{d}{dx}(x^3 e^x) - \frac{d}{dx}(2x)$

$f'(x) = (3x^2 \cdot e^x + x^3 \cdot e^x) - 2$

$f'(x) = e^x(x^3 + 3x^2) - 2$

3. Find the Second Derivative

We differentiate once more to find $f''(x)$. Again, we use the Product Rule on the term $e^x(x^3 + 3x^2)$:

$f''(x) = \frac{d}{dx}[e^x(x^3 + 3x^2)] - \frac{d}{dx}(2)$

$f''(x) = e^x(x^3 + 3x^2) + e^x(3x^2 + 6x) - 0$

4. Simplify the Expression

Factor out $e^x$ and combine the polynomial terms:

$f''(x) = e^x [x^3 + 3x^2 + 3x^2 + 6x]$

$f''(x) = e^x [x^3 + 6x^2 + 6x]$

To match the options provided in the image, we factor out a common $x$ from the bracket:

$f''(x) = x e^x (x^2 + 6x + 6)$

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Comparing our result to the given choices:

The correct option is 3.