Factorization of a^3+b^3+c^3-3abc

Stub explanation

Let $P(a,b,c)$ denote the three-variable polynomial $a^3+b^3+c^3-3abc$ (a homogeneous polynomial with "total" degree equal to $3$). It turns out we have the factorization $$P(a,b,c) = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^{-1}+c\omega^{-2}),$$ where $\omega = e^{2\pi i/3}$ is a primitive third root of unity. You can think of this as one possible extension of the difference of squares factorization $$a^2 - b^2 = (a+b)(a+b\zeta),$$ where $\zeta = e^{2\pi i/2} = -1$ is a primitive second root of unity (the more familiar extension being the sum and difference of two cubes).

To see why this is the case, one can expand and simplify to show $P(-b-c,b,c) = 0$. By the factor theorem for multivariable polynomials (reviewed here), this means $a - (-b-c) = a+b+c$ is a divisor/factor of $P(a,b,c)$. In fact, it turns out that $P(-b\omega-c\omega^2,b,c) = 0$ and $P(-b\omega^{-1}-c\omega^{-2},b,c) = 0$ as well, so $a+b\omega+c\omega^2$ and $a+b\omega^{-1}+c\omega^{-2}$ are also factors of $P(a,b,c)$.

Since $P$ is a homogeneous cubic, we deduce that $$P(a,b,c) = Z(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^{-1}+c\omega^{-2})$$ for some complex constant $Z\in\mathbb{C}$. In fact, $Z = 1$, as can be seen either by comparing---in the spirit of the method of undetermined coefficients---the coefficients of the $a^3$ terms (for instance), or by plugging in some choice of special values $(a,b,c) = (0,1,2)$ with $P(a,b,c) \ne 0$, and then solving for $Z$.

On the other hand, the irreducible factorization over the real numbers is $\left(a+b+c\right)\left(a^2+b^2+c^2-\left(ab+ac+bc\right)\right)$.

The factored form of a polynomial in the form of $a^3 + b^3 + c^3 - 3abc$ is $(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)$

Example 1:

$$x^3 + 8y^3 + 27z^3 - 18xyz$$

Note: $a = x$, $b = 2y$ and $c = 3z$

$$= (x + 2y + 3z)(x^2 + 4y^2 + 9z^2 - 2xy - 6yz - 3xz)$$

Example 2:

$$16x^3 + 54x^3y^3 + 250y^3z^3 - 180x^2y^2z$$ $$=2(8x^3 + 27x^3y^3 + 125y^3z^3)$$ $$=2(2x + 3xy + 5yz)(4x^2 + 9x^2y^2 + 25y^2z^2 - 6x^2y - 15xy^2z - 10xyz)$$

Test Yourself!

It turns out that the expression $a^3 + b^3 + c^3 - 3abc$ has the following factorization: $$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$$ The most common way of noticing this factorization is to try to make the polynomial equal to $0$ by placing a constraint on the values of $a, b, c$. In general, a constraint that makes a polynomial equal to $0$ corresponds to a factor of the polynomial. To this end, one might try to place a constraint on $a$ in terms of $b$ and $c$ to make the polynomial equal to $0$; it turns out that $a = -b-c$ does the trick nicely. Thus, $a+b+c=0$ implies that the polynomial equals $0$, so $a+b+c$ is a factor of the polynomial. By trying to find a polynomial to multiply with $a+b+c$ to obtain $a^3 + b^3 + c^3 - 3abc$, one can find the other factor without too much difficulty.

An interesting application of this fact is a proof of the AM-GM Inequality for $3$ variables. The AM-GM Inequality states that the arithmetic mean of a set of nonnegative real numbers is greater than or equal to the geometric mean of the same set of nonnegative real numbers, with equality if and only if all of the numbers are equal.

In the particular case of a set of $3$ nonnegative real numbers $x, y, z,$ the inequality becomes: $$\frac{1}{3}(x+y+z) \ge (x+y+z)^{1/3}$$ with equality if and only if $x=y=z$. Replacing $x, y, z$ with $a^3, b^3, c^3,$ respectively, and rearranging gives: $$a^3 + b^3 + c^3 - 3abc \ge 0$$ with equality if and only if $a = b = c$. The replacement is allowed because every nonnegative real number has a unique nonnegative real cube root.

Using the factorization, we can prove this inequality very quickly. We wish to prove: $$(a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)\ge 0$$ Since $a, b, c$ are nonnegative by the constraints of the inequality, $a+b+c\ge 0$. Now, notice that: $$a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2}\left[(a-b)^2 + (b-c)^2 + (c-a)^2\right]$$ which is at least $0$, since it is half of a sum of perfect squares. Therefore, both factors are nonnegative, so their product is nonnegative.

In the equality case, we have $a+b+c = 0$ or $a^2 + b^2 + c^2 - ab - bc - ca = 0$. The first one occurs if and only if $a=b=c=0$. The second one occurs if and only if all $3$ perfect square terms are $0$, meaning that $a-b=b-c=c-a=0$. This occurs if and only if $a=b=c$. Thus, equality holds if and only if $a=b=c$, meaning that all of the numbers are equal.

We have proved both the inequality and the equality case conditions, so we are done.

In the video above by the YouTube channel Kadas Learning it is explained how to factorize a polynomial of the form $a^3 + b^3 + c^3 - 3abc$

Determine all functions $f:\mathbb{Z}\to\mathbb{Z}$ such that, for every integer $a$, $f(a^3)=f(a)^3$ and for every triple $a,b,c$ of integers summing to $0$, we have $$f(a^3)+f(b^3)+f(c^3)=3f(a)f(b)f(c).$$