For all real numbers x and y such that x \neq y and x \neq -y, which of the following is e… — GRE General Test

For all real numbers $x$ and $y$ such that $x \neq y$ and $x \neq -y$ , which of the following is equivalent to the algebraic expression $\frac{x^3(x + 2y) - y^3(y + 2x)}{x^2 - y^2}$ ?

A
$x^2 + y^2$
$(x + y)^2$ Answer
C
$(x - y)^2$
D
$x^2 + 2xy - y^2$
E
$x + y$

Answer

The simplified expression is

(x + y)^2

Expanding and grouping the terms in the numerator gives

(x^4 - y^4) + 2xy(x^2 - y^2) = (x^2 - y^2)(x^2 + 2xy + y^2) = (x^2 - y^2)(x + y)^2

. Canceling the common factor

(x^2 - y^2)

from both the numerator and the denominator leaves

(x + y)^2

Step-by-Step Solution

Expand the numerator terms

x^3(x + 2y) - y^3(y + 2x) = x^4 + 2x^3y - y^4 - 2xy^3

Apply the distributive property to remove parentheses in the numerator.

Group the terms in the numerator to factor

(x^4 - y^4) + (2x^3y - 2xy^3) = (x^2 - y^2)(x^2 + y^2) + 2xy(x^2 - y^2)

Use difference of squares on

x^4 - y^4

and factor out the greatest common factor

2xy

from the remaining terms.

Factor out the common term

(x^2 - y^2)

from the numerator

(x^2 - y^2)(x^2 + y^2 + 2xy) = (x^2 - y^2)(x + y)^2

Recognize that

x^2 + 2xy + y^2

is the perfect square binomial

(x + y)^2

Simplify the rational expression by canceling common factors

\frac{(x^2 - y^2)(x + y)^2}{x^2 - y^2} = (x + y)^2

Divide numerator and denominator by

(x^2 - y^2)

, which is non-zero since

x \neq \pm y

Key Concept

Factoring high-degree algebraic expressions by grouping terms, recognizing difference of squares, and applying perfect square binomial identities.

Estimated Time:2m 0s

Question

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