Find the value

Let $[x / 2+y+z / 3]=a,[x / 3-2 y / 3+z]=b,[-5 x / 6-y / 3-4 z / 3]=c$

$a+b+c=x / 2+y+z / 3+x / 3-2 y / 3+z-5 x / 6-y / 3-4 z / 3$

$a+b+c=(x / 2+x / 3-5 x / 6)+(y-2 y / 3-y / 3)+(z / 3+z-4 z / 3)$

$a+b+c=3 x / 6+2 x / 6-5 x / 6+3 y / 3-2 y / 3-y / 3+z / 3+3 z / 3-4 z / 3$

$a+b+c=\frac{5 x-5 x}{6}+\frac{3 y-3 y}{3}+\frac{4 z-4 z}{3}$

$a+b+c=0$

$\because a+b+c=0 \therefore a^{3}+b^{3}+c^{3}=3 a b c$

$=3(x / 2+y+z / 3)(x / 3-2 y / 3+z)(-5 x / 6-y / 3-4 z / 3)$

$\therefore\left[x^{2}+y+z^{3}\right]^{3}+\left[x^{3}-2 y^{3}+z\right]^{3}+[-5 x / 6-y / 3-4 z / 3]^{3}$

$=3(x / 2+y+z / 3)(x / 3-2 y / 3+z)(-5 x / 6-y / 3-4 z / 3)$