Given that $\frac{49^{2x-3}}{7^{3x}} = \frac{7^{2x+1}}{343^{x-1}}$, find the value of $\sqrt[3]{343^x}$.

1. Express all terms in the equation with a common base of 7

$\frac{(7^2)^{2x-3}}{7^{3x}} = \frac{7^{2x+1}}{(7^3)^{x-1}}$

2. Apply laws of indices to simplify both sides of the equation

$\frac{7^{4x-6}}{7^{3x}} = \frac{7^{2x+1}}{7^{3x-3}}$
$7^{(4x-6) - 3x} = 7^{(2x+1) - (3x-3)}$
$7^{x-6} = 7^{-x+4}$

3. Equate the indices and solve for $x$

$x - 6 = -x + 4$
$2x = 10$
$x = 5$

4. Evaluate the required expression using the value of $x$

$\sqrt[3]{343^x} = \sqrt[3]{(7^3)^5}$
$= \sqrt[3]{7^{15}}$
$= 7^5$
$= 16807$