Let $A = \begin{bmatrix} x & 1 \\ 0 & 1 \end{bmatrix}$ where $x$ is a real number, and $B = \begin{bmatrix} 2\sqrt{2} & 3 + \sqrt{2} \\ 0 & 1 \end{bmatrix}$. If $A^3 = B$, then $x$ is equal to

Since $A^3 = B$, their determinants must also be equal.

Also, We know that $|A^3| = |A|^3$.

Now, finding the determinant of $A$:

$|A| = \begin{vmatrix} x & 1 \\ 0 & 1 \end{vmatrix}$ $= x(1) - 0(1) = x$

Therefore, $|A^3| = |A|^3 = x^3$

Next, finding the determinant of $B$:

$|B| = \begin{vmatrix} 2\sqrt{2} & 3+\sqrt{2} \\ 0 & 1 \end{vmatrix}$

$= 2\sqrt{2}(1) - 0(3+\sqrt{2}) = 2\sqrt{2}$

Since $|A^3| = |B|$,

$x^3 = 2\sqrt{2} = (\sqrt{2})^3$

Hence, $x = \sqrt{2}$.

The answer is Option 2: $\sqrt{2}$