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Overview

Many olympiad algebra problems involve expressions of degree three, four, or higher. These expressions initially can seem impossible to simplify.

However, many higher degree expressions possess hidden structure. By recognising standard factorisations and important identities, we can often transform a complicated expression into something much more manageable.

These ideas appear a lot in AMC, AIME and olympiad algebra.

Motivation

Suppose we wish to factor

x6−1.x^6-1.

Expanding is clearly not a sensible approach.

Instead notice that

x6−1=(x3)2−(1)2.x^6-1=(x^3)^2-(1)^2.

This immediately becomes a difference of squares:

x6−1=(x3−1)(x3+1).x^6-1=(x^3-1)(x^3+1).

Each factor can then be factored again using the cubic identities:

(x−1)(x2+x+1)(x+1)(x2−x+1).(x-1)(x^2+x+1)(x+1)(x^2-x+1).

Rather than performing tedious computations, we can use structure to split the expression apart.

This is the main idea of higher power factorisation.

Difference of Powers

One of the most important factorisation formulas is

an−bn.a^n-b^n.

For every positive integer nn,

an−bn=(a−b)(an−1+an−2b+⋯+abn−2+bn−1).a^n-b^n = (a-b) (a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}).

The factor (a−b)(a-b) always appears.

Example

Factor

x5−1.x^5-1.

Applying the formula gives

x5−1=(x−1)(x4+x3+x2+x+1).x^5-1 = (x-1) (x^4+x^3+x^2+x+1).

This factorisation appears frequently in number theory and polynomial problems.

Sum of Powers (Odd Exponent)

When nn is odd, we also have a factorisation for an+bna^n + b^n:

an+bn=(a+b)(an−1−an−2b+an−3b2−⋯+bn−1).a^n+b^n = (a+b) (a^{n-1}-a^{n-2}b+a^{n-3}b^2-\cdots+b^{n-1}).

The signs in the second factor alternate.

The sum of cubes (n=3n=3) is the most common special case, but n=5n=5 and n=7n=7 also appear in olympiad problems.

Note that when nn is even, an+bna^n + b^n does not factor over the integers in general.

Difference of Squares

The most fundamental factorisation is

a2−b2=(a−b)(a+b).a^2-b^2 = (a-b)(a+b).

Many higher power factorisations begin by repeatedly applying this identity.

Example

Factor

x8−y8.x^8-y^8.

Apply difference of squares repeatedly:

x8−y8=(x4−y4)(x4+y4)x^8-y^8 = (x^4-y^4)(x^4+y^4)

and

x4−y4=(x2−y2)(x2+y2)x^4-y^4 = (x^2-y^2)(x^2+y^2)

and

x2−y2=(x−y)(x+y).x^2-y^2 = (x-y)(x+y).

As such

x8−y8=(x−y)(x+y)(x2+y2)(x4+y4).x^8-y^8 = (x-y)(x+y)(x^2+y^2)(x^4+y^4).

Difference and Sum of Cubes

Two of the most useful factorisations are

Difference of Cubes

a3−b3=(a−b)(a2+ab+b2).a^3-b^3 = (a-b)(a^2+ab+b^2).

Sum of Cubes

a3+b3=(a+b)(a2−ab+b2).a^3+b^3 = (a+b)(a^2-ab+b^2).

These are special cases of the difference and sum of powers formulas above, and they appear throughout olympiad algebra.

Example

Factor

x3+8.x^3+8.

Since

8=23,8=2^3,

we get

x3+8=(x+2)(x2−2x+4).x^3+8 = (x+2)(x^2-2x+4).

Example

Factor

x6−y6.x^6-y^6.

Using

x6−y6=(x3−y3)(x3+y3),x^6-y^6=(x^3-y^3)(x^3+y^3),

we get

(x−y)(x2+xy+y2)(x+y)(x2−xy+y2).(x-y)(x^2+xy+y^2) (x+y)(x^2-xy+y^2).

Fourth Powers

Expressions involving fourth powers often hide simpler structure.

For example,

a4−b4=(a2−b2)(a2+b2).a^4-b^4 = (a^2-b^2)(a^2+b^2).

Applying difference of squares again gives

a4−b4=(a−b)(a+b)(a2+b2).a^4-b^4 = (a-b)(a+b)(a^2+b^2).

This factorisation appears frequently in divisibility and polynomial problems.

The Symmetric Cubic Identity

One of the most important olympiad identities is

a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca).a^3+b^3+c^3-3abc = (a+b+c) (a^2+b^2+c^2-ab-bc-ca).

This is known as the symmetric cubic identity.

The second factor can also be written as

12[(a−b)2+(b−c)2+(c−a)2],\frac{1}{2} \left[ (a-b)^2+(b-c)^2+(c-a)^2 \right],

which shows that for real numbers, the second factor is always non negative. This means the sign of a3+b3+c3−3abca^3+b^3+c^3-3abc is determined entirely by the sign of a+b+ca+b+c (when a,b,ca,b,c are real and not all equal).

Proof

Consider

(a+b+c)(a2+b2+c2−ab−bc−ca).(a+b+c) (a^2+b^2+c^2-ab-bc-ca).

Expanding and collecting like terms leads to

a3+b3+c3−3abc.a^3+b^3+c^3-3abc.

Although the expansion is somewhat lengthy, it is completely straightforward.

The important fact to remember is the resulting identity rather than the expansion itself.

The Special Case a+b+c=0a+b+c=0

A particularly useful consequence occurs when

a+b+c=0.a+b+c=0.

Substituting into the symmetric cubic identity gives

a3+b3+c3−3abc=0.a^3+b^3+c^3-3abc=0.

Therefore

a3+b3+c3=3abc.a^3+b^3+c^3 = 3abc.

This is one of the most frequently used identities in olympiad algebra. Whenever a problem gives or implies that three quantities sum to zero, this identity should come to mind.

Worked Examples

Example 1

Given that a+b+c=0a+b+c=0, evaluate

a3+b3+c3a^3+b^3+c^3.

Using the special case above,

a3+b3+c3=3abc.a^3+b^3+c^3 = 3abc.

No expansion is required.

Example 2

Factor x3+y3+z3−3xyzx^3+y^3+z^3-3xyz.

Applying the symmetric cubic identity gives

(x+y+z)(x2+y2+z2−xy−yz−zx).(x+y+z) (x^2+y^2+z^2-xy-yz-zx).

Example 3

Given x+y+z=0x+y+z=0, simplify

x3+y3+z3+5xyzx^3+y^3+z^3+5xyz.

Since

x3+y3+z3=3xyz,x^3+y^3+z^3 = 3xyz,

the expression becomes

3xyz+5xyz=8xyz.3xyz+5xyz = 8xyz.

Example 4

Given a+b+c=0a+b+c=0, simplify

a3+b3+c3abc.\frac{a^3+b^3+c^3}{abc}.

Since a+b+c=0a+b+c=0, we have a3+b3+c3=3abca^3+b^3+c^3 = 3abc, so

a3+b3+c3abc=3abcabc=3\frac{a^3+b^3+c^3}{abc} = \frac{3abc}{abc} = 3

Example 5

Show that n5−nn^5 - n is divisible by 3030 for all integers nn.

Factor using difference of powers:

n5−n=n(n4−1)=n(n2−1)(n2+1).n^5 - n = n(n^4 - 1) = n(n^2 - 1)(n^2 + 1).

Continuing:

=n(n−1)(n+1)(n2+1).= n(n-1)(n+1)(n^2+1).

Among any three consecutive integers n−1,n,n+1n-1, n, n+1, one is divisible by 22 and one by 33, so n5−nn^5 - n is divisible by 66.

For divisibility by 55, check all residues modulo 55. If n≡0n \equiv 0 then n≡0n \equiv 0. If n≡1n \equiv 1 then n−1≡0n-1 \equiv 0. If n≡4n \equiv 4 then n+1≡0n+1 \equiv 0. If n≡2n \equiv 2 then n2+1≡0n^2+1 \equiv 0. If n≡3n \equiv 3 then n2+1≡0n^2+1 \equiv 0.

In every case, one of the factors is divisible by 55. Since gcd⁡(5,6)=1\gcd(5,6)=1, we conclude 30∣n5−n30 \mid n^5 - n.

Key Contest Insight

When working with high degree expressions, avoid expanding immediately.

Instead look for:

  • difference of squares,
  • sum or difference of cubes (or higher odd powers),
  • repeated factorisations,
  • factors of the form a+b+ca+b+c,
  • and opportunities to apply a3+b3+c3−3abca^3+b^3+c^3-3abc.

Many olympiad problems are designed so that recognising the correct identity is the hardest step.

Strategy Tips

  • Always search for common factorisation patterns before expanding.
  • If an expression contains a+b+ca+b+c, check whether the symmetric cubic identity may apply.
  • Repeatedly factor differences of squares whenever possible.
  • Look for opportunities to rewrite exponents:
x8−y8=(x4)2−(y4)2.x^8-y^8=(x^4)^2-(y^4)^2.
  • If you are given a+b+c=0a+b+c=0, consider
a3+b3+c3=3abc.a^3+b^3+c^3=3abc.

Common Pitfalls

  • Expanding large expressions unnecessarily.
  • Forgetting the sign in the sum of cubes formula.
  • Applying the symmetric cubic identity incorrectly.
  • Missing repeated applications of difference of squares.
  • Forgetting that
a3+b3+c3=3abca^3+b^3+c^3=3abc

requires the condition

a+b+c=0.a+b+c=0.

Practice Problems

StatusSourceProblem NameDifficultyTags
AIMEMedium
Show TagsHigher Power Factorization, Sum of Cubes, Vieta's Formulas
AMC 12Medium
Show TagsDifference of Squares, Higher Power Factorization, Telescoping
AMC 10Hard
Show TagsBinary Representation, Higher Power Factorization, Sum of Powers
Berkeley Math CircleHard
Show TagsFactorization techniques, Techniques: modulo, size analysis, order analysis, inequalities
Berkeley Math CircleHard
Show TagsChinese remainder theorem, Factorization techniques
Berkeley Math Circle Monthly Contest 7Hard
Show TagsFactorization techniques, Greatest common divisors (gcd)
MathNetHard
Show TagsComplex numbers, Factorization techniques
Berkeley Math Circle Monthly Contest 3Hard
Show TagsFactorization techniques, τ (number of divisors)
Berkeley Math Circle: Monthly Contest 8Hard
Show TagsFactorization techniques, Polynomial operations
Berkeley Math Circle Monthly Contest 1Hard
Show TagsFactorization techniques, Fermat / Euler / Wilson theorems, Integers, Techniques: modulo, size analysis, order analysis, inequalities
Berkeley Math CircleHard
Show TagsFactorization techniques
Berkeley Math Circle Monthly Contest 6Hard
Show TagsFactorization techniques, Integers
Berkeley Math Circle Monthly Contest 3Hard
Show TagsFactorization techniques, Polynomial operations, Prime numbers
Berkeley Math CircleHard
Show TagsFactorization techniques, Techniques: modulo, size analysis, order analysis, inequalities
Berkeley Math Circle Monthly Contest 6Hard
Show TagsFactorization techniques
Berkeley Math Circle Monthly Contest 3Hard
Show TagsFactorization techniques, Multiplicative order
Berkeley Math Circle: Monthly Contest 6Hard
Show TagsFactorization techniques, Floors and ceilings
Berkeley Math CircleHard
Show TagsCounting two ways, Factorization techniques, τ (number of divisors)
Berkeley Math Circle Monthly Contest 5Hard
Show TagsFactorization techniques, Polynomial operations, Sums and products, Techniques: modulo, size analysis, order analysis, inequalities
Berkeley Math Circle Monthly Contest 4Hard
Show TagsFactorization techniques, Integers, Polynomial operations
Berkeley Math Circle: Monthly Contest 1Hard
Show TagsFactorization techniques
Berkeley Math Circle Monthly Contest 7Hard
Show TagsFactorization techniques, Integers
MathNetHard
Show TagsFactorization techniques, Polynomial operations
Berkeley Math Circle Monthly Contest 4Hard
Show TagsFactorization techniques, Induction / smoothing

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