Factorization of a polynomials in complex number.

Question

Factorize this expression:

$$a^2+b^2+c^2-ab-bc-ca.$$

The result is

$$(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)$$

How I can get $\Omega$ here?What's the approach?

Answer 1

we have $$(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)$$$$=a^2+ab\Omega^2+ac\Omega+ab\Omega+b^2\Omega^3+bc\Omega^2+ac\Omega^2+bc\Omega^4+c^2\Omega^3$$ $$=a^2+\Omega^3b^2+\Omega^3c^2+(\Omega^2+\Omega)ab+(\Omega^2+\Omega^4)bc+(\Omega+\Omega^2)ac$$ $$=a^2+b^2+c^2-ab-bc-ac$$ It follows that $$\Omega^3=1$$ and $$\Omega^2+\Omega=\Omega^2+\Omega^4=-1.$$ Solving $$\Omega^2+\Omega=-1$$ will give you two solutions that work (check that they satisfy all of the above equations). These happen to be the primitive third roots of unity as $$\Omega^2+\Omega+1$$ is the third cyclotomic polynomial.

Answer 2

This is equal to: $$\frac{a^3+b^3+c^3-3abc}{a+b+c}$$ which you can verify by multiplying your polynomial by $a+b+c$. So let's try to factor $a^3+b^3+c^3-3abc$.

So we know that $$a+b+c\mid a^3+b^3+c^3-3abc$$ (The vertical line means "is a factor of.")

Now, let $\Omega$ be a cube root of unity (usually denoted by $\omega$ but i'll follow the notation in your question). Notice that the maps: \begin{align} a&\mapsto a\\ b&\mapsto b\Omega\\ c&\mapsto c\Omega^2\\ \end{align} don't change $a^3+b^3+c^3-3abc$. Thus, applying them to the previous equation, we get: \begin{align} a+b+c&\mid a^3+b^3+c^3-3abc\\ a+b\Omega+c\Omega^2&\mid a^3+b^3+c^3-3abc\\ a+b\Omega^2+c\Omega&\mid a^3+b^3+c^3-3abc \end{align} These are its only factors, since it's a monic degree-3 polynomial and we've just supplied three monic degree-1 factors. Thus: $$(a+b+c)(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)\\=a^3+b^3+c^3-3abc$$ Dividing out by $a+b+c$ and using the observation at the start of this answer, we get your factorization.

Answer 3

If we are given that

$a^2 + b^2 + c^2 - ab - ac - bc$ $ = (a + b\Omega + c\Omega^2)(a + b\Omega^2 + c\Omega), \tag{1}$

and we wish to find $\Omega$, we can proceed as follows: multiplying out the right-hand side, we find that

$a^2 + b^2 + c^2 - ab - ac - bc$ $ = a^2 + \Omega^3 b^2 + \Omega^3 c^2 $ $+ (\Omega^2 + \Omega)ab + (\Omega^2 + \Omega)ac + (\Omega^4 + \Omega^2) bc; \tag{2}$

we see from (2) that the factorization succeeds provided that

$\Omega^3 = 1, \tag{3}$

$\Omega^2 + \Omega = -1, \tag{4}$

and

$\Omega^4 + \Omega^2 = -1; \tag{5}$

it follows from (3) that $\Omega$ must be one of the three cube roots of unity $1$, $e^{2\pi /3}$, $e^{4\pi i /3}$; it follows from (4), (5) that $\Omega \ne 1$; thus, since

$(\Omega - 1)(\Omega^2 + \Omega + 1) = \Omega^3 - 1 = 0, \tag{6}$

(4) must bind; the apparent discrepancy 'twixt (4) and (5) is resolved by observing that (3) yields

$\Omega^4 = \Omega \Omega^3 = \Omega(1) = \Omega; \tag{7}$

(4) and (5) are in reality the same equation. We thus see that the factorization (1) holds with $\Omega = e^{2 \pi i / 3}$, $\Omega = e^{4 \pi i / 3}$, since both satisfy (3)-(5).

Answer 4

Let's do it as follows for a change.

Consider the circulant matrix $$ M(a,b,c):=\left(\begin{array}{ccc}a&b&c\\c&a&b\\b&c&a\end{array}\right) =a I+ b T+ c T^2, $$ where $T=M(0,1,0)$ satisfies the relations $T^2=M(0,0,1), T^3=M(1,0,0)=I$.

Let's calculate $\det M(a,b,c)$. Let's first add the 2nd and 3rd rows to the first. The first row then becomes a constant $a+b+c$. Therefore $$ \det M(a,b,c)=(a+b+c)\left\vert\begin{array}{ccc}1&1&1\\c&a&b\\b&c&a\end{array}\right\vert=(a+b+c)\left[(a^2-bc)+(b^2-ac)+(c^2-ab)\right], $$ IOW apart from the factor $a+b+c$ we see that your polynomial is the determinant of $M(a,b,c)$.

But the eigenvalues of $T$ are $\lambda_1=1$, $\lambda_2=\omega:=e^{2\pi/3}$ and $\lambda_3=\omega^2$ - the third roots of unity. Therefore the eigenvalues of $M(a,b,c)$ are $a\lambda_i^0+b\lambda_i^1+c\lambda_i^2, i=1,2,3.$ These evaluate to $$ a+b+c,\quad a+\omega b+\omega^2c,\quad a+\omega^2 b+\omega c. $$ The determinant of a matrix is, of course, the product of its eigenvalues. Cancelling the known extra factor $a+b+c$ gives you what you want.

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This is, of course, overkill for the purposes of factoring this quadratic, but the method generalizes (once you recognize what is required), and circulant matrices are nice to chat about.