26. Investigate the possibility of equality:

$$
k y^{3}=x^{3}+z^{3}
$$

where $k$ is an integer, if $x, y, z$ are integers forming an increasing arithmetic progression.
26. From the system $\left\{\begin{array}{l}x^{3}+z^{3}=k y^{3} \\ x+z=2 y\end{array}\right.$ we find $k y^{3}=2 y\left(x^{2}-\right.$ $\left.-x z+z^{2}\right) ;$ therefore, 1) $\left.y=0, x=-z ; 2\right)(k-8) x^{2}+$ $+2(k+4) x z+(k-8) z^{2}=0, \frac{x}{z}=\frac{-(k+4) \pm \sqrt{24(k-2)}}{k-8}$, if $k \neq 8, k-2=6 t^{2}$; thus, $k$ must be of the form $2\left(3 t^{2}+1\right)$, where $t$ is any integer; therefore, $\left(\frac{x}{z}\right)_{1}=\frac{1-t}{1+t},\left(\frac{x}{z}\right)_{2}=\frac{1+t}{1-t}$, from which we find $x=(1-t) m, \quad y=m, \quad z=(1+t) m ; \quad x=(1+t) m, \quad y=m$, $z=(1-t) m$.

By the condition $x, y, z$ form an increasing arithmetic progression, so we can take $x=(1-$ $-t) m, y=m, z=(1+t) m$, where $t$ and $m$ are any positive integers. Consider $k=8$.