For all real numbers x and y such that x \neq y, x \neq -y, and x^2 + y^2 \neq 0, which of… — GRE General Test

For all real numbers

x

and

y

such that

x \neq y

x \neq -y

, and

x^2 + y^2 \neq 0

, which of the following expressions is equivalent to

\frac{x^4 - y^4}{x^3 - x^2 y + x y^2 - y^3} - \frac{2 x y}{x + y}

\frac{x^2 + y^2}{x + y}

Factoring

x^4 - y^4

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

and

x^3 - x^2 y + x y^2 - y^3

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

simplifies the first term to

x + y

. Combining

x + y

with

-\frac{2xy}{x+y}

over the common denominator

(x + y)

gives

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

Factor the numerator and denominator of the first rational expression

Numerator:

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

. Denominator:

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

Factoring allows for cancellation of common factors in rational expressions.

Simplify the first rational expression by canceling common terms

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

Since

x \neq y

and

x^2 + y^2 \neq 0

, the common terms

(x - y)

and

(x^2 + y^2)

are non-zero and can be divided out.

Subtract the second expression using a common denominator

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

Combining terms under the common denominator

(x + y)

enables algebraic reduction.

Expand the squared binomial in the numerator and simplify like terms

\frac{x^2 + 2xy + y^2 - 2xy}{x + y} = \frac{x^2 + y^2}{x + y}.

Expanding

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

allows the

+2xy

and

-2xy

terms to cancel out.

Simplifying rational expressions using polynomial factoring (difference of squares and grouping) and common denominators.

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