# Trouble on Factorizing

I have trouble factoring $$a^2+ab+b^2$$. It can be done easily using $$\omega$$, which is a complex cube root of unity with non-zero imaginary part- $$a^2+ab+b^2 = a^2 - \omega ab - {\omega}^2ab+{\omega}^3b^2 = a(a-\omega b) -\omega ^2b(a-\omega b) = (a-\omega ^2b)(a-\omega b)$$ But the following is also a way to factor- $$a^2+ab+b^2 = (a+b)^2-ab = (a+b-\sqrt {ab})(a+b+\sqrt {ab})$$ But why is it that this factorization not so much popular while the one with complex factors is more popular? Can anybody explain me with an example?
What I think is it is probably because $$\sqrt{ab}$$ can't be real all the time. Thanks!

• Because it involves radicals? And complex numbers usually make calculation easier Jul 24, 2020 at 13:31
• Because $\sqrt{ab}$ is not a polynomial. When talking about factoring something, it should normally be clear (at least from context if not stated explicitly) what kind of factors are desired. When the object to be factored is a polynomial, one would generally expect that this is talking about factoring into polynomials. Jul 24, 2020 at 14:03
• So is $\omega$ is a constituent of a polynomial? (real coefficients) Jul 24, 2020 at 14:13
• @Book Of Flames, it is a coefficient, but it is not a real coefficient. Jul 24, 2020 at 15:49
• @UmbQbify-Key20-, So the coefficients need not be real every time, right? Jul 24, 2020 at 15:57

The first factorisation is a product of two linear factors, and it can immediately be seen that the two zeroes of the axpression are for $$a=\omega b$$ and $$a=\omega^2 b$$.
The second factorisation contains a term in $$\sqrt{ab}$$, and there is no such simple expression to find the zeroes. In fact finding a value for $$\frac{a}{b}$$ to give a zero for either factor, would effectively involve going back to the original expression.