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Difficulty: MediumSimplifying and Factoring Algebraic Expressions

For all real numbers xx and yy such that y≠0y \neq 0 and 3x2+y2≠03x^2 + y^2 \neq 0, which of the following expressions is equivalent to (x+y)3−(x−y)32y(3x2+y2)\frac{(x + y)^3 - (x - y)^3}{2y(3x^2 + y^2)}?

  1. 11Answer
  2. B
    y23x2+y2\frac{y^2}{3x^2 + y^2}
  3. C
    3x23x2+y2\frac{3x^2}{3x^2 + y^2}
  4. D
    3x+y3x2+y2\frac{3x + y}{3x^2 + y^2}
  5. E
    13\frac{1}{3}

Answer

11
Expanding the numerator yields (x3+3x2y+3xy2+y3)−(x3−3x2y+3xy2−y3)=6x2y+2y3=2y(3x2+y2)(x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3) = 6x^2y + 2y^3 = 2y(3x^2 + y^2). Since the numerator and denominator are identical non-zero expressions, the quotient simplifies to 1.

Step-by-Step Solution

1
Expand (x+y)3(x + y)^3 and (x−y)3(x - y)^3 using the binomial theorem.
(x+y)3=x3+3x2y+3xy2+y3(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 and (x−y)3=x3−3x2y+3xy2−y3(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3
Expanding the binomial expressions allows combination of like terms in the numerator.
2
Subtract (x−y)3(x - y)^3 from (x+y)3(x + y)^3.
(x^3 + 3x^2y + 3xy^2 + y^3) - (x^3 - 3x^2y + 3xy^2 - y^3) = 6x^2y + 2y^3
The terms x3x^3 and 3xy23xy^2 subtract to zero, while the remaining terms double.
3
Factor out common terms from the simplified numerator.
6x2y+2y3=2y(3x2+y2)6x^2y + 2y^3 = 2y(3x^2 + y^2)
Factoring out 2y2y reveals a factor identical to the denominator.
4
Divide the numerator by the denominator 2y(3x2+y2)2y(3x^2 + y^2).
2y(3x2+y2)2y(3x2+y2)=1\frac{2y(3x^2 + y^2)}{2y(3x^2 + y^2)} = 1
Since y≠0y \neq 0 and 3x2+y2≠03x^2 + y^2 \neq 0, identical non-zero factors cancel out.

Key Concept

Binomial expansion of cubic expressions and factoring algebraic expressions.
Estimated Time:1m 30s
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Topics

Simplifying and Factoring Algebraic Expressions

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