If a, b, and c are pairwise distinct real numbers, which of the following expressions is e… — GRE General Test

a

b

, and

c

are pairwise distinct real numbers, which of the following expressions is equivalent to

\frac{a^3(b - c) + b^3(c - a) + c^3(a - b)}{(a - b)(b - c)(c - a)}?

-(a + b + c)

Factoring the numerator by using the Factor Theorem and cyclic symmetry reveals that

a^3(b - c) + b^3(c - a) + c^3(a - b) = -(a - b)(b - c)(c - a)(a + b + c)

. Dividing this by the denominator

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

cancels the pairwise difference terms, leaving

-(a + b + c)

Analyze the numerator for polynomial factors using cyclic symmetry

Let

P(a, b, c) = a^3(b - c) + b^3(c - a) + c^3(a - b)

. If

a = b

, then

P(b, b, c) = b^3(b - c) + b^3(c - b) + 0 = 0

. By the Factor Theorem,

(a - b)

is a factor. By cyclic symmetry,

(b - c)

and

(c - a)

are also factors.

Identifying linear factors reduces the polynomial simplification problem.

Determine the degree and form of the remaining factor

P(a, b, c)

is a homogeneous polynomial of degree 4, while

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

has degree 3. Therefore, the remaining factor must be a homogeneous symmetric polynomial of degree 1, which takes the form

k(a + b + c)

for some constant

k

Homogeneous degree properties dictate the algebraic structure of the quotient.

Find the constant

k

by substituting test values

Let

a = 0

b = 1

, and

c = 2

. Evaluating

P(0, 1, 2) = 0 + 1^3(2 - 0) + 2^3(0 - 1) = 2 - 8 = -6

. The factor product gives

(0 - 1)(1 - 2)(2 - 0) = (-1)(-1)(2) = 2

. Setting

2 \cdot k(0 + 1 + 2) = -6 \implies 6k = -6 \implies k = -1

Evaluating at convenient integer values determines the missing constant scalar.

Divide the factored numerator by the denominator

\frac{-(a - b)(b - c)(c - a)(a + b + c)}{(a - b)(b - c)(c - a)} = -(a + b + c)

Canceling common non-zero factors yields the simplified expression.

Factoring Cyclic Symmetric Polynomials

Question