If \(x=a-b, y=b-c, z=\mathrm{c}-a\), then the numerical value of the algebraic expression \(x^{3}+y^{3}+z^{3}-3 x y z\) will be

To find the numerical value of the given algebraic expression \(x^{3}+y^{3}+z^{3}-3xyz\) when \(x=a-b\), \(y=b-c\), and \(z=c-a\), we can substitute these values directly into the expression. However, there’s a known identity in algebra that can simplify our efforts:

\[
x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)\left(x^{2}+y^{2}+z^{2}-xy-xz-yz\right)
\]

First, let’s find \(x+y+z\):

\[
x+y+z = (a-b)+(b-c)+(c-a)
\]

Simplifying this, we find:

\[
x+y+z = 0
\]

Since \(x+y+z = 0\), the whole expression \(x^{3}+y^{3}+z^{3}-3xyz\) simplifies to 0 because anything multiplied by 0 is 0. Therefore, the numerical value of the given algebraic expression is \(0\).

The correct answer is (b) 0.