Expanding and Factoring

Algebraic manipulation is the bread and butter of olympiad algebra. While you likely know the basic identities, success in competitions relies on recognizing these patterns in complex, disguised forms.

Fundamental Identities

Theorem (Standard Expansions)

For any complex numbers

a, b, c

Square of Sum: $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (ab + b c + c a)$
Difference of Squares: $a^{2} - b^{2} = (a - b) (a + b)$
Difference of Cubes: $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$
Sum of Cubes: $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$

Advanced Identities

1. Sophie Germain's Identity

This identity is frequently used in number theory and algebra problems involving terms of the form $a^{4} + 4 b^{4}$ .

Theorem (Sophie Germain Identity)

a^{4} + 4 b^{4} = (a^{2} + 2 b^{2} + 2 ab) (a^{2} + 2 b^{2} - 2 ab)

Proof. Add and subtract $4 a^{2} b^{2}$ to complete the square:

a^{4} + 4 b^{4} = a^{4} + 4 a^{2} b^{2} + 4 b^{4} - 4 a^{2} b^{2} = (a^{2} + 2 b^{2})^{2} - (2 ab)^{2} = (a^{2} + 2 b^{2} - 2 ab) (a^{2} + 2 b^{2} + 2 ab)

∎

2. Euler's Identity

Theorem (Euler's Factorization)

a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)

Note

A crucial consequence is that if

a + b + c = 0

, then

a^{3} + b^{3} + c^{3} = 3 ab c

. This "conditional identity" appears often in competition problems.

Olympiad Techniques

1. The "Add Zero" Trick

Sometimes an expression becomes factorable only after adding and subtracting a specific term (like in the proof of Sophie Germain's Identity).

Example

Factor

x^{4} + x^{2} + 1

Proof (Solution). We want to complete the square for $(x^{2} + 1)^{2} = x^{4} + 2 x^{2} + 1$ .

x^{4} + x^{2} + 1 = (x^{4} + 2 x^{2} + 1) - x^{2} = (x^{2} + 1)^{2} - x^{2} = (x^{2} + 1 - x) (x^{2} + 1 + x)

∎

2. Simon's Favorite Factoring Trick (SFFT)

Used to handle expressions like

x y + x + y

Tip (SFFT)

x y + x + y + 1 = (x + 1) (y + 1)

Generally, to factor

A x y + B x + C y

, we multiply by

A

and add

BC

A^{2} x y + A B x + A C y + BC = (A x + C) (A y + B)

Practice Problems

Exercise (Problem 1)

Factor completely:

x^{4} - 20 x^{2} + 4

Exercise (Problem 2)

Given

a + b + c = 0

, prove that

2 (a^{4} + b^{4} + c^{4}) = (a^{2} + b^{2} + c^{2})^{2}

Exercise (Problem 3)

Compute the value of

\frac{202 3 ^{3} + 1}{202 3 ^{2} - 2022}