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Difficulty: HardSimplifying and Factoring Algebraic Expressions

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

P(x,y)=(x3+y3−x2y−xy2x4−y4)⋅(x3+y3x2−xy+y2)P(x, y) = \left(\frac{x^3 + y^3 - x^2 y - xy^2}{x^4 - y^4}\right) \cdot \left(\frac{x^3 + y^3}{x^2 - xy + y^2}\right)

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

  1. x4−y4(x2+y2)2\frac{x^4 - y^4}{(x^2 + y^2)^2}Answer
  2. 1−2y2x2+y21 - \frac{2y^2}{x^2 + y^2}Answer
  3. C
    (x−y)2x2+y2\frac{(x - y)^2}{x^2 + y^2}
  4. D
    x−yx+y\frac{x - y}{x + y}
  5. E
    x2+(−y)2x2−y2\frac{x^2 + (-y)^2}{x^2 - y^2}

Answer

The expressions equivalent to P(x,y)P(x, y) are x4−y4(x2+y2)2\frac{x^4 - y^4}{(x^2 + y^2)^2} and 1−2y2x2+y21 - \frac{2y^2}{x^2 + y^2}.
The given expression simplifies to x2−y2x2+y2\frac{x^2 - y^2}{x^2 + y^2}. The option x4−y4(x2+y2)2\frac{x^4 - y^4}{(x^2 + y^2)^2} simplifies directly to x2−y2x2+y2\frac{x^2 - y^2}{x^2 + y^2} after factoring the numerator. The option 1−2y2x2+y21 - \frac{2y^2}{x^2 + y^2} simplifies to x2+y2−2y2x2+y2=x2−y2x2+y2\frac{x^2 + y^2 - 2y^2}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2} when combined over a common denominator.

Step-by-Step Solution

1
Factor the numerator of the first rational term by grouping terms.
x3+y3−x2y−xy2=x2(x−y)−y2(x−y)=(x2−y2)(x−y)=(x−y)2(x+y)x^3 + y^3 - x^2 y - xy^2 = x^2(x - y) - y^2(x - y) = (x^2 - y^2)(x - y) = (x - y)^2 (x + y)
Grouping allows rewriting four polynomial terms into product of linear/quadratic factors.
2
Factor the denominator of the first rational term as a difference of squares.
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 x4−y4x^4 - y^4 into simpler factors reveals common factors with the numerator.
3
Simplify the first rational term by canceling common factors (x−y)(x+y)(x - y)(x + y).
\frac{(x - y)^2(x + y)}{(x - y)(x + y)(x^2 + y^2)} = \frac{x - y}{x^2 + y^2}
Canceling non-zero common factors simplifies the fraction.
4
Factor the numerator of the second rational term using the sum of cubes formula.
x3+y3=(x+y)(x2−xy+y2)x^3 + y^3 = (x + y)(x^2 - xy + y^2)
Applying the sum of cubes identity exposes the irreducible quadratic factor present in the denominator.
5
Simplify the second rational term and multiply the result by the simplified first term.
P(x,y)=(x−yx2+y2)⋅(x+y)=(x−y)(x+y)x2+y2=x2−y2x2+y2P(x, y) = \left(\frac{x - y}{x^2 + y^2}\right) \cdot (x + y) = \frac{(x - y)(x + y)}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}
Multiplying the simplified forms yields the simplest explicit representation of P(x,y)P(x, y).
6
Verify equivalence of the options against x2−y2x2+y2\frac{x^2 - y^2}{x^2 + y^2}.
The option x4−y4(x2+y2)2=(x2−y2)(x2+y2)(x2+y2)2=x2−y2x2+y2\frac{x^4 - y^4}{(x^2 + y^2)^2} = \frac{(x^2 - y^2)(x^2 + y^2)}{(x^2 + y^2)^2} = \frac{x^2 - y^2}{x^2 + y^2}, and the option 1−2y2x2+y2=x2+y2−2y2x2+y2=x2−y2x2+y21 - \frac{2y^2}{x^2 + y^2} = \frac{x^2 + y^2 - 2y^2}{x^2 + y^2} = \frac{x^2 - y^2}{x^2 + y^2}. Both match P(x,y)P(x, y).
Transforming algebraic expressions under common denominators or factoring confirms equivalence.

Key Concept

Simplifying complex algebraic expressions using factoring by grouping, difference of squares, sum of cubes, and common denominator manipulation.
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Topics

Simplifying and Factoring Algebraic Expressions

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