For all real numbers x and y such that 4x^2 + 6xy + 9y^2 \neq 0 and 2x + 3y \neq 0, which … — GRE General Test

For all real numbers $x$ and $y$ such that $4x^2 + 6xy + 9y^2 \neq 0$ and $2x + 3y \neq 0$ , which of the following expressions are equivalent to $\frac{8x^3 - 27y^3}{4x^2 + 6xy + 9y^2}$ ? Select all such expressions.

$2x - 3y$ Answer
$-(3y - 2x)$ Answer
$\frac{4x^2 - 9y^2}{2x + 3y}$ Answer
D
$2x + 3y$
E
$(2x - 3y)^2$

Answer

The equivalent expressions are

2x - 3y

-(3y - 2x)

, and

\frac{4x^2 - 9y^2}{2x + 3y}

Factoring

8x^3 - 27y^3

(2x - 3y)(4x^2 + 6xy + 9y^2)

allows canceling the denominator, leaving

2x - 3y

. The expression

2x - 3y

is directly correct. Rearranging terms in

-(3y - 2x)

yields

-3y + 2x = 2x - 3y

, making it correct. Factoring

\frac{4x^2 - 9y^2}{2x + 3y}

\frac{(2x - 3y)(2x + 3y)}{2x + 3y}

simplifies to

2x - 3y

, making it also correct.

Step-by-Step Solution

Factor the numerator of the rational expression using the difference of cubes formula.

8x^3 - 27y^3 = (2x)^3 - (3y)^3 = (2x - 3y)((2x)^2 + (2x)(3y) + (3y)^2) = (2x - 3y)(4x^2 + 6xy + 9y^2)

Recognizing

8x^3

(2x)^3

and

27y^3

(3y)^3

enables full polynomial factoring.

Simplify the fraction by canceling the common non-zero quadratic factor.

\frac{(2x - 3y)(4x^2 + 6xy + 9y^2)}{4x^2 + 6xy + 9y^2} = 2x - 3y

Since

4x^2 + 6xy + 9y^2 \neq 0

, dividing common factors reduces the expression to linear form.

Evaluate each provided choice for algebraic equivalence to

2x - 3y

The expressions

2x - 3y

-(3y - 2x) = 2x - 3y

, and

\frac{4x^2 - 9y^2}{2x + 3y} = \frac{(2x-3y)(2x+3y)}{2x+3y} = 2x - 3y

are all algebraically identical.

Re-expressing or factoring alternative choices proves their equivalence to the simplified expression.

Key Concept

Factoring difference of cubes

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

and difference of squares

a^2 - b^2 = (a - b)(a + b)

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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