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Difficulty: Very hardSimplifying and Factoring Algebraic Expressions

For all real numbers xx and yy such that x2+xy+y2≠0x^2 + xy + y^2 \neq 0, which of the following expressions are equivalent to x6−y6x2+xy+y2\frac{x^6 - y^6}{x^2 + xy + y^2}? Select all such expressions.

  1. (x2−y2)(x2−xy+y2)(x^2 - y^2)(x^2 - xy + y^2)Answer
  2. (x−y)(x3+y3)(x - y)(x^3 + y^3)Answer
  3. x4−x3y+xy3−y4x^4 - x^3 y + x y^3 - y^4Answer
  4. D
    (x+y)(x3−y3)(x + y)(x^3 - y^3)
  5. E
    x4−y4x^4 - y^4

Answer

The equivalent expressions are (x2−y2)(x2−xy+y2)(x^2 - y^2)(x^2 - xy + y^2), (x−y)(x3+y3)(x - y)(x^3 + y^3), and x4−x3y+xy3−y4x^4 - x^3y + xy^3 - y^4.
Factoring the numerator x6−y6x^6 - y^6 as a difference of squares yields (x3−y3)(x3+y3)(x^3 - y^3)(x^3 + y^3). Applying the difference of cubes identity x3−y3=(x−y)(x2+xy+y2)x^3 - y^3 = (x - y)(x^2 + xy + y^2) enables cancellation of the non-zero denominator x2+xy+y2x^2 + xy + y^2, leaving (x−y)(x3+y3)(x - y)(x^3 + y^3). Expanding this product gives x4−x3y+xy3−y4x^4 - x^3y + xy^3 - y^4. Furthermore, factoring x3+y3x^3 + y^3 as (x+y)(x2−xy+y2)(x + y)(x^2 - xy + y^2) and regrouping (x−y)(x+y)(x - y)(x + y) gives (x2−y2)(x2−xy+y2)(x^2 - y^2)(x^2 - xy + y^2). Consequently, the three valid equivalent forms are (x2−y2)(x2−xy+y2)(x^2 - y^2)(x^2 - xy + y^2), (x−y)(x3+y3)(x - y)(x^3 + y^3), and x4−x3y+xy3−y4x^4 - x^3y + xy^3 - y^4.

Step-by-Step Solution

1
Factor the numerator x6−y6x^6 - y^6 as a difference of squares.
x6−y6=(x3)2−(y3)2=(x3−y3)(x3+y3)x^6 - y^6 = (x^3)^2 - (y^3)^2 = (x^3 - y^3)(x^3 + y^3)
Applying the difference of squares identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b) with a=x3a = x^3 and b=y3b = y^3.
2
Apply the difference of cubes identity to x3−y3x^3 - y^3.
x3−y3=(x−y)(x2+xy+y2)x^3 - y^3 = (x - y)(x^2 + xy + y^2)
This reveals the non-zero quadratic factor present in the denominator.
3
Substitute into the original fraction and cancel the common factor (x2+xy+y2)(x^2 + xy + y^2).
(x−y)(x2+xy+y2)(x3+y3)x2+xy+y2=(x−y)(x3+y3)\frac{(x - y)(x^2 + xy + y^2)(x^3 + y^3)}{x^2 + xy + y^2} = (x - y)(x^3 + y^3)
Since x2+xy+y2≠0x^2 + xy + y^2 \neq 0, dividing numerator and denominator by (x2+xy+y2)(x^2 + xy + y^2) simplifies the expression to (x−y)(x3+y3)(x - y)(x^3 + y^3).
4
Expand (x−y)(x3+y3)(x - y)(x^3 + y^3) to check for equivalent expanded polynomial forms.
(x−y)(x3+y3)=x4+xy3−x3y−y4=x4−x3y+xy3−y4(x - y)(x^3 + y^3) = x^4 + xy^3 - x^3y - y^4 = x^4 - x^3y + xy^3 - y^4
Distributing terms verifies polynomial equivalence.
5
Factor x3+y3x^3 + y^3 and regroup to find another equivalent factored representation.
(x−y)(x3+y3)=(x−y)(x+y)(x2−xy+y2)=(x2−y2)(x2−xy+y2)(x - y)(x^3 + y^3) = (x - y)(x + y)(x^2 - xy + y^2) = (x^2 - y^2)(x^2 - xy + y^2)
Applying the sum of cubes identity x3+y3=(x+y)(x2−xy+y2)x^3 + y^3 = (x + y)(x^2 - xy + y^2) and combining (x−y)(x+y)=x2−y2(x - y)(x + y) = x^2 - y^2.

Key Concept

Factoring higher-degree algebraic expressions using difference of squares and sum/difference of cubes identities
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Topics

Simplifying and Factoring Algebraic Expressions

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