Cyclic and Symmetric Expressions

In the Chinese book: 轮换式与对称式

Letters swap seats but the expression stays put. One factor gives you the rest.

Try this

Click “Rotate”: $a$ becomes $b$ , $b$ becomes $c$ and $c$ becomes $a$ . The expression is written differently, but after expanding and simplifying it is exactly the same as before. An expression like this is called cyclic.
Reset, then click “Swap a↔b”: this time every term changes sign. So it is not symmetric. (An expression is symmetric if swapping any two letters leaves it unchanged.)
Click “Try $a - b$ ”: setting $a = b$ makes the whole expression $0$ . That shows $(a - b)$ is a factor.
Since rotating leaves the expression unchanged, the rotated copies of $(a - b)$ , namely $(b - c)$ and $(c - a)$ , must be factors too. The original has degree $3$ , and the three linear factors multiply to degree $3$ as well, so they differ only by a constant. The workbench finds it is $- 1$ .

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

That is what symmetry is good for: find one factor by testing, and the others need no testing: just rotate.

Predict first

Of the three expressions $a + b + c$ , $ab + b c + c a$ and $a^{2} b + b^{2} c + c^{2} a$ , how many are symmetric?

1

2

3

Choose an answer first; then you can see the result.

Why

Why does “setting $a = b$ gives $0$ ” mean $(a - b)$ is a factor? Think of the expression as a polynomial in $a$ alone, with $b$ and $c$ treated as numbers. This is exactly the Factor Theorem from stop 8: substituting $a = b$ gives $0$ , so $(a - b)$ is a factor.

Example: factor a³ + b³ + c³ − 3abc

Step 1 / 5

Look at the symmetry first
Swap $a$ and $b$ : the expression is unchanged. The same goes for any other two letters. It is a symmetric expression.

$b^{3} + a^{3} + c^{3} - 3 ba c$
So rotating changes nothing either
Try a − b
Try a + b + c
The quotient is symmetric too

The routine for factoring symmetric expressions: taken together, the factors of a symmetric expression must be symmetric too. So there are very few candidates: try the $(a - b)$ kind first, then the $(a + b)$ kind, then $(a + b + c)$ . Once the factors are fixed, count the degrees, and substitute one set of numbers to find any remaining coefficient.

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: $(a - b)$ is a factor $\Rightarrow$ $(b - c)$ and $(c - a)$ are factors too

$a^{2} (b - c) + b^{2} (c - a) + c^{2} (a - b)$ is cyclic. Once $(a - b)$ passes the test, the other two come along automatically.

Practice

Exercise 1Basic

Swap $a$ and $b$ in $(a - b) (b - c) (c - a)$ . The result is the original multiplied by what?

Exercise 2Advanced

Factor $ab (a + b) + b c (b + c) + c a (c + a) + 2 ab c$ .

Exercise 3Advanced

Factor $ab (a - b) + b c (b - c) + c a (c - a)$ .

Exercise 4Challenge

Factor $a^{3} (b - c) + b^{3} (c - a) + c^{3} (a - b)$ .

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Mistaking a cyclic expression for a symmetric one

a^{2} b + b^{2} c + c^{2} a

the three letters play the same role, so it is symmetric, and swapping

a

and

b

leaves it unchanged.

What went wrong? Think first, then open

Diagnosis Swapping $a$ and $b$ gives $b^{2} a + a^{2} c + c^{2} b$ , which is not the same expression (take $a = 1, b = 2, c = 0$ : the original is $2$ , the swapped one is $4$ ). It is unchanged only under the rotation $a \to b \to c \to a$ .

A symmetric expression is unchanged when any two letters are swapped; a cyclic expression only needs to be unchanged when the three letters are rotated in turn. Every symmetric expression is cyclic, but not the other way round.

“Symmetric ⇒ cyclic, but cyclic ⇏ symmetric”

Exhibit

Stopping before the degrees match

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

: setting

a = b

gives

0

, so it equals

k (a - b) (b - c) (c - a)

What went wrong? Think first, then open

Diagnosis The original has degree $4$ , while $(a - b) (b - c) (c - a)$ has degree only $3$ . A linear factor is still missing; the two cannot differ by just a constant.

The missing linear factor must also be unchanged by rotation, so it can only be

k (a + b + c)

. Substitute one set of numbers to find

k = - 1

: the original

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

“Count the degrees first, then pin down the remaining factor”