How do you factor completely \[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz\]?

Hint: In this problem we have to completely factor the given cubic equation. To solve these types of problems, we have to know some basic algebraic formulas. In this problem we are going to use the sum of cubes formula in the first two terms, to factor the given equation, after applying the cubes formula we will take some common terms and apply algebraic formulas to get the factor of the equation.

Complete step by step answer:
We know that the given equation is,
\[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz\] …… (1)
We know that the sum of cubes formula is,
\[{{x}^{3}}+{{y}^{3}}={{\left( x+y \right)}^{3}}-3xy\left( x+y \right)\].
Now we can apply the above formula in the equation (1), we get
\[\Rightarrow {{\left( x+y \right)}^{3}}-3xy\left( x+y \right)+{{z}^{3}}-3xyz\]
Now we can take the cube terms first and then we can write the remaining terms,
\[\Rightarrow {{\left( x+y \right)}^{3}}+{{z}^{3}}-3xy\left( x+y \right)-3xyz\]
We can now take the first two cube terms and apply the sum of the cubes identity.
\[\begin{align}
  & \Rightarrow {{\left[ \left( x+y \right)+z \right]}^{3}}-3\left( x+y \right)z\left[ \left( x+y \right)+z \right]-3xy\left( x+y \right)-3xyz \\
 & \Rightarrow {{\left( x+y+z \right)}^{3}}-3\left( x+y \right)z\left( x+y+z \right)-3xy\left( x+y+z \right) \\
\end{align}\]
Now we can take the common term \[x+y+z\]from the above step, we get
\[\Rightarrow \left( x+y+z \right)\left[ {{\left( x+y+z \right)}^{2}}-3xz-3yz-3xy \right]\] ……. (3)
We also know that the algebraic identity,
\[{{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx\]
Now we can apply the above identity in the expression (3), we get
\[\begin{align}
  & \Rightarrow \left( x+y+z \right)\left[ \left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx \right)-3xy-3yz-3zx \right] \\
 & \Rightarrow \left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2zx-3xy-3yz-3zx \right) \\
\end{align}\]
Now we can subtract the similar terms, we get
\[\Rightarrow \left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)\]
Therefore, the factor of \[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz\] is \[\left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)\].

Note:
Students make mistakes in writing the algebraic identities, which should be concentrated to solve or factorize these types of problems. We should also know to apply the correct identity or formula in the suitable step. Students also make mistakes while taking the common terms outside, which must be concentrated.