Polynomials in Multiple Variables

When dealing with 3 or more variables, symmetry plays a massive role. Most Olympiad problems in this category involve symmetric sums or cyclic shifts.

Symmetric Polynomials

A polynomial $P (x_{1}, \dots, x_{n})$ is symmetric if permuting the variables does not change the polynomial. Examples:

$x + y + z$
$x y + yz + z x$
$x^{3} + y^{3} + z^{3}$

Fundamental Theorem of Symmetric Polynomials

Any symmetric polynomial in

x_{1}, \dots, x_{n}

can be expressed uniquely as a polynomial in the elementary symmetric polynomials:

$e_{1} = \sum x_{i}$
$e_{2} = \sum x_{i} x_{j}$
$e_{3} = \sum x_{i} x_{j} x_{k}$
...
$e_{n} = x_{1} x_{2} \dots x_{n}$

Cyclic Polynomials

A polynomial is cyclic if it remains unchanged under the cyclic shift

(x_{1} \to x_{2} \to \dots \to x_{n} \to x_{1})

Note: All symmetric polynomials are cyclic, but not all cyclic polynomials are symmetric (e.g., $x^{2} y + y^{2} z + z^{2} x$ ).

Factorization Tricks

1. Vanishing Check

If setting

x = y

makes the polynomial zero, then

(x - y)

is a factor. If the polynomial is symmetric/cyclic, then

(y - z)

and

(z - x)

must also be factors.

Example

Factor

P (x, y, z) = x (y - z)^{3} + y (z - x)^{3} + z (x - y)^{3}

Proof. Set $x = y$ : $y (y - z)^{3} + y (z - y)^{3} + z (0) = y (y - z)^{3} - y (y - z)^{3} = 0$ . So $(x - y)$ is a factor. By cyclic symmetry, $(y - z)$ and $(z - x)$ are factors. Since $de g (P) = 4$ , we must have: $P (x, y, z) = K (x + y + z) (x - y) (y - z) (z - x)$ . Comparing coefficients (e.g., $x y^{3}$ term) gives $K = 1$ . ∎

Practice Problems

Exercise (Problem 1)

Factor completely:

(a + b + c)^{3} - (a^{3} + b^{3} + c^{3})

Exercise (Problem 2)

Given

x + y + z = 0

, express

x^{4} + y^{4} + z^{4}

in terms of

S = x^{2} + y^{2} + z^{2}

Exercise (Problem 3)

Let

a, b, c

be roots of

P (x) = x^{3} - x + 1

. Compute

a^{5} + b^{5} + c^{5}

without finding the roots explicitly.