Factorize each of the following expressions:$ \left(\frac{x}{2}+y+\frac{z}{3}\right)^{3}+\left(\frac{x}{3}-\frac{2 y}{3}+z\right)^{3} +\left(-\frac{5 x}{6}-\frac{y}{3}-\frac{4 z}{3}\right)^{3} $

Given:

\( \left(\frac{x}{2}+y+\frac{z}{3}\right)^{3}+\left(\frac{x}{3}-\frac{2 y}{3}+z\right)^{3}  +\left(-\frac{5 x}{6}-\frac{y}{3}-\frac{4 z}{3}\right)^{3} \)

To do:

We have to multiply the given expressions.

Solution:

We know that,

$a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$

$a^3 + b^3 + c^3 = 3abc$ if $a + b + c = 0$

Here,

$\frac{x}{2}+y+\frac{z}{3}+\frac{x}{3}-\frac{2 y}{3}+z+\frac{-5 x}{6}-\frac{y}{3}-\frac{4}{3} z=\frac{x}{2}+\frac{x}{3}-\frac{5 x}{6}+y-\frac{2 y}{3}-\frac{y}{3}+\frac{z}{3}+z-\frac{4}{3} z$

$=\frac{3 x+2 x-5 x}{6}+\frac{3 y-2 y-y}{3}+\frac{z+3 z-4 z}{3}$

$=0+0+0$

$=0$

Therefore,

$(\frac{x}{2}+y+\frac{z}{3})^{3}+(\frac{x}{3}-\frac{2 y}{3}+z)^{3}  +(-\frac{5 x}{6}-\frac{y}{3}-\frac{4 z}{3})^{3} = 3 (\frac{x}{2}+y+\frac{z}{3})(\frac{x}{3}-\frac{2 y}{3}+z)(-\frac{5 x}{6}-\frac{y}{3}-\frac{4 z}{3})$

Hence, $(\frac{x}{2}+y+\frac{z}{3})^{3}+(\frac{x}{3}-\frac{2 y}{3}+z)^{3}  +(-\frac{5 x}{6}-\frac{y}{3}-\frac{4 z}{3})^{3} = 3 (\frac{x}{2}+y+\frac{z}{3})(\frac{x}{3}-\frac{2 y}{3}+z)(-\frac{5 x}{6}-\frac{y}{3}-\frac{4 z}{3})$.