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

For all real numbers

x

and

y

such that

|x| \neq |y|

, which of the following expressions is equivalent to

\frac{x^3 + x^2y + x^2 - y^3 - xy^2 - y^2}{x^2 - y^2}?

x + y + 1

Factoring the numerator by grouping yields

(x - y)(x + y)(x + y + 1)

. Factoring the denominator gives

(x - y)(x + y)

. Canceling the non-zero common factor

(x - y)(x + y)

leaves the simplified expression

x + y + 1

Group terms in the numerator to identify common factor pairs.

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

Grouping cubic terms, quadratic cross-terms, and difference of squares separately allows factoring out fundamental algebraic patterns.

Apply standard algebraic formulas to each grouped term.

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

Using difference of cubes

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

and difference of squares

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

reveals a common

(x-y)

factor across all terms.

Factor out

(x - y)

from the numerator and simplify the remaining polynomial.

N = (x - y)\left[(x^2 + xy + y^2) + xy + (x + y)\right] = (x - y)\left[x^2 + 2xy + y^2 + x + y\right]

Combining like terms inside the bracket simplifies the expression.

Recognize the perfect square trinomial inside the expression.

N = (x - y)\left[(x + y)^2 + (x + y)\right] = (x - y)(x + y)(x + y + 1)

Rewriting

x^2 + 2xy + y^2

(x + y)^2

allows factoring out

(x + y)

Divide the factored numerator by the denominator.

\frac{(x - y)(x + y)(x + y + 1)}{(x - y)(x + y)} = x + y + 1

Since

|x| \neq |y|

, both

(x - y)

and

(x + y)

are non-zero and can be canceled.

Factoring multivariable polynomials using grouping, difference of cubes, and difference of squares formulas.

Soru