CDS Mathematics · Algebraic Identities and Simplification

Cyclic Sums and Factors

Sums that cycle a → b → c → a have fixed values and fixed factors, and a single numerical test is the fastest way to find them.

All Algebraic Identities and Simplification notes CDS Mathematics strategy

Why this matters

Six PYQs, every one of them HARD, and all six are faster by testing numbers than by algebra. Three sums are worth knowing by heart, and one principle: if an expression vanishes when two letters are equal, their difference is a factor.

Concept 1 of 2: Three cyclic fraction sums

Fractions with denominators

(a - b)(a - c)

(b - c)(b - a)

(c - a)(c - b)

add up to a polynomial of small degree, often just a constant. Put in

1, 2, 3

and read the answer.

Definition

For distinct $a, b, c$ , summing over the three cyclic terms:

$\displaystyle\sum \dfrac{1}{(a - b)(a - c)} = 0$
$\displaystyle\sum \dfrac{a}{(a - b)(a - c)} = 0$
$\displaystyle\sum \dfrac{a^2}{(a - b)(a - c)} = 1$
$\displaystyle\sum \dfrac{a^3}{(a - b)(a - c)} = a + b + c$

Note $(b - a)(c - a) = (a - b)(a - c)$ , but $(b - a)(a - c)$ is the negative. Rewrite every denominator in the standard order before using these.

The key cyclic sum

\dfrac{a^2}{(a - b)(a - c)} + \dfrac{b^2}{(b - c)(b - a)} + \dfrac{c^2}{(c - a)(c - b)} = 1

Worked example

Check

\displaystyle\sum \dfrac{a}{(a - b)(a - c)} = 0

a, b, c = 1, 2, 3

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2023 · CDS (II) 2023 — Elementary Mathematics · Q74Hard

Example 1 · Algebraic Identities and Simplification · Symmetric and Cyclic Expressions

p = \frac{a^2}{(b - a)(c - a)}

q = \frac{b^2}{(c - b)(a - b)}

r = \frac{c^2}{(a - c)(b - c)}

, then what is

(p + q + r)^2

equal to ?

Watch the order of the factors

Swapping one factor,

(a - b)

for

(b - a)

, flips the sign of that term. A sum that 'should' be

1

can be

-1

if the question writes the denominators differently — check with numbers.

Drill 3 more on three cyclic fraction sums

Concept 2 of 2: Factors of cyclic expressions

If an expression becomes zero when

a = b

, it has

(a - b)

as a factor. A cyclic expression then has

(b - c)

and

(c - a)

too. Compare degrees to see what is left, and test one set of numbers to fix the constant and the sign.

Definition

$a(b^2 - c^2) + b(c^2 - a^2) + c(a^2 - b^2) = (a - b)(b - c)(c - a)$ .
If the expression has degree $n$ , the remaining factor has degree $n - 3$ and is itself symmetric or cyclic.
To test whether $(a + b + c)$ is a factor, evaluate at a distinct triple with $a + b + c = 0$ , such as $(1, 2, -3)$ : a non-zero value proves it is not.
The sign of $(x - y)(y - z)(z - x)$ is negative when $x > y > z$ .

The basic cyclic factorisation

a(b^2 - c^2) + b(c^2 - a^2) + c(a^2 - b^2) = (a - b)(b - c)(c - a)

Worked example

Check the basic factorisation at

(a, b, c) = (0, 1, 2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2022 · CDS (II) 2022 — Elementary Mathematics · Q11Hard

Example 2 · Algebraic Identities and Simplification · Symmetric and Cyclic Expressions

Consider the following statements : 1.

x^4(y - z) + y^4(z - x) + z^4(x - y)

is positive if

x > y > z

. 2.

x^4(y - z) + y^4(z - x) + z^4(x - y)

is negative if

x < y < z

. Which of the above statements is/are correct ?

One test disproves; it does not prove

A single triple giving a non-zero value shows a factor is absent. A triple giving zero shows nothing by itself — use the vanishing-when-equal argument to prove a factor is present.

Drill 1 more on factors of cyclic expressions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Three cyclic fraction sums
The key cyclic sum
$\dfrac{a^2}{(a - b)(a - c)} + \dfrac{b^2}{(b - c)(b - a)} + \dfrac{c^2}{(c - a)(c - b)} = 1$
Factors of cyclic expressions
The basic cyclic factorisation
$a(b^2 - c^2) + b(c^2 - a^2) + c(a^2 - b^2) = (a - b)(b - c)(c - a)$

Watch out for (2)

Watch the order of the factors
→ Three cyclic fraction sums
One test disproves; it does not prove
→ Factors of cyclic expressions

Test yourself on Algebraic Identities and Simplification

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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