Question

Difficulty: MediumSimplifying and Factoring Algebraic Expressions

For all real numbers xx and yy such that x≠yx \neq y, x≠−yx \neq -y, and x2+y2≠0x^2 + y^2 \neq 0, consider the algebraic expression:

E(x,y)=x4−y4x3−x2y+xy2−y3E(x, y) = \frac{x^4 - y^4}{x^3 - x^2y + xy^2 - y^3}

Which of the following expressions are equivalent to E(x,y)E(x, y) for all valid values of xx and yy? Select all such expressions.

  1. x2−y2x−y\frac{x^2 - y^2}{x - y}Answer
  2. x3+y3x2−xy+y2\frac{x^3 + y^3}{x^2 - xy + y^2}Answer
  3. C
    x2+y2x+y\frac{x^2 + y^2}{x + y}
  4. D
    x3−y3x2+xy+y2\frac{x^3 - y^3}{x^2 + xy + y^2}
  5. E
    x2−y2x+y\frac{x^2 - y^2}{x + y}

Answer

The expressions equivalent to E(x,y)E(x, y) are x2−y2x−y\frac{x^2 - y^2}{x - y} and x3+y3x2−xy+y2\frac{x^3 + y^3}{x^2 - xy + y^2}.
Simplifying E(x,y)E(x, y) by factoring both numerator and denominator yields x+yx + y. The option with x2−y2x−y\frac{x^2 - y^2}{x - y} simplifies directly to x+yx + y by canceling (x−y)(x - y). The option with x3+y3x2−xy+y2\frac{x^3 + y^3}{x^2 - xy + y^2} uses the sum of cubes identity to factor the numerator into (x+y)(x2−xy+y2)(x + y)(x^2 - xy + y^2), which also cancels down to x+yx + y. Both of these options are mathematically identical to E(x,y)E(x, y).

Step-by-Step Solution

1
Factor the numerator of E(x,y)E(x, y) using the difference of squares identity twice.
x4−y4=(x2−y2)(x2+y2)=(x−y)(x+y)(x2+y2)x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2)
Decomposing higher-power binomials into linear and quadratic factors enables algebraic simplification.
2
Factor the denominator of E(x,y)E(x, y) by grouping terms.
x3−x2y+xy2−y3=x2(x−y)+y2(x−y)=(x−y)(x2+y2)x^3 - x^2y + xy^2 - y^3 = x^2(x - y) + y^2(x - y) = (x - y)(x^2 + y^2)
Grouping adjacent terms with shared factors allows factoring out (x−y)(x - y).
3
Simplify the full expression E(x,y)E(x, y) by canceling common non-zero factors.
E(x,y)=(x−y)(x+y)(x2+y2)(x−y)(x2+y2)=x+yE(x, y) = \frac{(x - y)(x + y)(x^2 + y^2)}{(x - y)(x^2 + y^2)} = x + y
Since x≠yx \neq y and x2+y2≠0x^2 + y^2 \neq 0, the factors (x−y)(x - y) and (x2+y2)(x^2 + y^2) cancel completely.
4
Evaluate the given options to determine which simplify to x+yx + y.
x2−y2x−y=x+y\frac{x^2 - y^2}{x - y} = x + y and x3+y3x2−xy+y2=x+y\frac{x^3 + y^3}{x^2 - xy + y^2} = x + y, whereas x3−y3x2+xy+y2=x−y\frac{x^3 - y^3}{x^2 + xy + y^2} = x - y and x2−y2x+y=x−y\frac{x^2 - y^2}{x + y} = x - y.
Matching each option's fully simplified form to x+yx + y identifies all valid equivalent expressions.

Key Concept

Simplifying rational expressions by polynomial factoring (grouping, difference of squares, and sum of cubes).
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