Simple. (Problem $4$ )

$\dfrac{x}{2^3} = x \quad \implies \quad \frac{x}{x} = 2^3 \quad [x \ne 0] \quad \implies \quad \because 1 \ne 8 \quad [x \ne 0] \text{ is not true } \quad \implies \quad \boxed{x = 0}$

$\frac{x}{2^{3}}=x$

$\frac{x}{8}=x$

$\frac{x}{8}-x=0$

$-\frac{7x}{8}=0$

$\Rightarrow$ $x=\boxed{0}$

Since $\frac{0}{y}$ will always be $0$ , no matter what $y$ is (apart from itself as $\frac{0}{0}$ $= 1$ ).

Therefore, $x = \fbox{0}$ .