5. Solution: Let $x=a+d, y=a+2d, z=a+3d, w=a+4d$. Then from
$$x^{3}+y^{3}+z^{3}=w^{3}$$

we get
$$(a+d)^{3}+(a+2d)^{3}+(a+3d)^{3}=(a+4d)^{3}$$

Expanding and combining like terms, we obtain
$$a^{3}+3a^{2}d-3ad^{2}-14d^{3}=0$$

This is $\square$
$$a^{3}-2a^{2}d+5a^{2}d-10ad^{2}+7ad^{2}-14d^{3}=0$$

Thus, we have $\square$
$$(a-2d)(a^{2}+5ad+7d^{2})=0$$

Since $25-4 \times 7=-30$

From (1), we get $a=2d$, so the solution is
$$x=3d, y=4d, z=5d, w=6d$$

where $d$ is any natural number.