For pairwise distinct real numbers x, y, and z, consider the algebraic expression: E(x, y,… — GRE General Test

For pairwise distinct real numbers $x$ , $y$ , and $z$ , consider the algebraic expression:

E(x, y, z) = \frac{(x^2 - y^2)^3 + (y^2 - z^2)^3 + (z^2 - x^2)^3}{(x - y)^3 + (y - z)^3 + (z - x)^3}

If $x = 5$ , $y = 3$ , and $z = 1$ , what is the numerical value of $E(5, 3, 1)$ ?

Answer: 192

192

Using the identity that

a^3 + b^3 + c^3 = 3abc

whenever

a + b + c = 0

, both the numerator and denominator can be factored directly. Factoring the difference of squares in the numerator yields

3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)

. Dividing this by the factored denominator

3(x - y)(y - z)(z - x)

simplifies the expression to

(x + y)(y + z)(z + x)

. Substituting

x = 5

y = 3

, and

z = 1

yields

(8)(4)(6) = 192

Use the conditional cubic identity

a + b + c = 0 \implies a^3 + b^3 + c^3 = 3abc

on the denominator.

Denominator becomes

3(x - y)(y - z)(z - x)

The sum of the three terms

(x - y) + (y - z) + (z - x)

equals 0.

Apply the same identity to the numerator.

Numerator becomes

3(x^2 - y^2)(y^2 - z^2)(z^2 - x^2)

The sum of the squared difference terms

(x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2)

also equals 0.

Factor each difference of squares in the numerator.

Numerator becomes

3(x - y)(x + y)(y - z)(y + z)(z - x)(z + x)

Using the difference of squares identity

u^2 - v^2 = (u - v)(u + v)

on each term.

Simplify the fraction by dividing the common factors in the numerator and denominator.

E(x, y, z) = (x + y)(y + z)(z + x)

The factors

3

(x - y)

(y - z)

, and

(z - x)

cancel out completely.

Evaluate the simplified product for

x = 5

y = 3

, and

z = 1

(5 + 3)(3 + 1)(1 + 5) = 8 \times 4 \times 6 = 192

Direct evaluation after algebraic simplification.

Simplifying rational expressions involving sum of cubes identity

a^3 + b^3 + c^3 = 3abc

when

a + b + c = 0

and difference of squares factoring.

Question