### Theory:

Identity $$VIII$$: ${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ac}\right)$

$$21x^3 + 8y^3 + z^3 - 12xyz$$

Let us write the expression as $$21x^3 + 8y^3 + z^3 - 12xyz$$.

Using the identity $$VIII$$: ${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ac}\right)$

${\phantom{\rule{0.147em}{0ex}}\mathit{21x}}^{3}+{\mathit{8y}}^{3}+{z}^{3}-12\mathit{xyz}=\left(3x+2y+z\right)\left({\mathit{9x}}^{2}+{\mathit{4y}}^{2}+{z}^{2}-\left(\mathit{3x}\right)\left(\mathit{2y}\right)-\left(\mathit{2y}\right)\left(z\right)-\left(3x\right)\left(z\right)\right)$

${\phantom{\rule{0.147em}{0ex}}\mathit{21x}}^{3}+{\mathit{8y}}^{3}+{z}^{3}-12\mathit{xyz}=\left(3x+2y+z\right)\left({\mathit{3x}}^{2}+{\mathit{2y}}^{2}+{z}^{2}-6\mathit{xy}-2\mathit{yz}-3\mathit{xz}\right)$

Important!
If ${a}^{3}+{b}^{3}+{c}^{3}=0$

Then, ${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ac}\right)$

We know that, ${a}^{3}+{b}^{3}+{c}^{3}=0$

${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(0\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ac}\right)$

$\begin{array}{l}{a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(0\right)\\ \\ {a}^{3}+{b}^{3}+{c}^{3}=3\mathit{abc}\end{array}$