A faster way than tedious multiplying

Question

Show that $(a+b+c)(a+b\epsilon +c\epsilon^2)(a+b\epsilon^2 + c\epsilon) = a^3 + b^3 + c^3 - 3abc$ If $$\epsilon^2 + \epsilon + 1 =0$$

The solution in the back of the book is given as

Proved by a direct check, taking into consideration that $\epsilon^2 = -\epsilon -1 \: \:$ and $\epsilon^3 = 1$

However the solutions still feels like tedious multiplication, doesn't it? Is there a faster, more elegant way to do this?

Answer 1

One way would be by factorizing $a^3+b^3+c^3-3abc$.

$$a^3+b^3+c^3-3abc=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$$ where $\omega$ is the cube root of unity.

More on the factorization can be found here.

Then, clearly, $\epsilon=\omega$, and thus, $$\epsilon^3=1$$ Factorizing this gives, $$(\epsilon-1)(\epsilon^2+\epsilon+1)=0$$ Thus, $$\epsilon^2+\epsilon+1=0$$