Factorizations Using Cube Roots of Unity

The symmetric factorization of a³+b³+c³−3abc over ℂ. When a+b+c = 0, the identity gives a³+b³+c³ = 3abc directly — a classic JEE trick. Also: a²+b²+c²−ab−bc−ca = (a+ωb+ω²c)(a+ω²b+ωc).

Over the reals, $a^3 + b^3 + c^3 - 3abc$ factors as $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. Over $\mathbb{C}$, using cube roots of unity, it splits completely into three linear factors.

The Complete Factorization

Claim:

$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a + \omega b + \omega^2 c)(a + \omega^2 b + \omega c)$$

Derivation

Treat the right side as a product and expand the last two factors first.

Let $P = (a + \omega b + \omega^2 c)(a + \omega^2 b + \omega c)$.

Expanding:

$$P = a^2 + a\omega^2 b + a\omega c + \omega ab + \omega^3 b^2 + \omega^2 bc + \omega^2 ac + \omega^4 bc + \omega^3 c^2$$

Using $\omega^3 = 1$ and $\omega^4 = \omega$:

$$P = a^2 + b^2 + c^2 + ab(\omega + \omega^2) + bc(\omega^2 + \omega) + ca(\omega + \omega^2)$$

Since $1 + \omega + \omega^2 = 0$, we have $\omega + \omega^2 = -1$:

$$P = a^2 + b^2 + c^2 - ab - bc - ca$$

Now multiply by $(a + b + c)$:

$$(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = a^3+b^3+c^3-3abc$$

(This is the standard real factorization identity.)

Therefore:

$$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a + \omega b + \omega^2 c)(a + \omega^2 b + \omega c) \qquad \blacksquare$$

The Key Corollary

If $a + b + c = 0$, then:

$$a^3 + b^3 + c^3 = 3abc$$

This is one of the most-used tricks in JEE algebra. Whenever you spot three quantities summing to zero, their cubes sum to three times their product.

The Related Identity

$$a^2 + b^2 + c^2 - ab - bc - ca = (a + \omega b + \omega^2 c)(a + \omega^2 b + \omega c)$$

This is exactly $P$ from above. It also equals $\dfrac{1}{2}\left[(a-b)^2 + (b-c)^2 + (c-a)^2\right]$, confirming it is always $\geq 0$ for real $a, b, c$, with equality iff $a = b = c$.