For all real numbers x such that x \neq 3 and x \neq -3, the algebraic expression \frac{\f… — GRE General Test

For all real numbers

x

such that

x \neq 3

and

x \neq -3

, the algebraic expression

\frac{\frac{x^4 - 81}{x^2 - 9} \cdot \frac{x^3 - 27}{x^2 + 3x + 9} - x(x - 3)^2}{x^2 - 9}

simplifies to a constant value. What is the value of this constant?

Answer: 3

The simplified expression evaluates to the constant value

3

Factoring the numerator components using difference of squares and difference of cubes simplifies the product term to

x^3 - 3x^2 + 9x - 27

. Subtracting

x(x - 3)^2 = x^3 - 6x^2 + 9x

simplifies the entire numerator to

3x^2 - 27 = 3(x^2 - 9)

. Dividing by the denominator

(x^2 - 9)

cancels out the variable terms entirely, yielding the constant value 3.

Simplify the first rational component using the difference of squares identity

\frac{x^4 - 81}{x^2 - 9} = \frac{(x^2 - 9)(x^2 + 9)}{x^2 - 9} = x^2 + 9

Since

x \neq \pm 3

x^2 - 9 \neq 0

, allowing direct cancellation of

(x^2 - 9)

Simplify the second rational component using the difference of cubes identity

\frac{x^3 - 27}{x^2 + 3x + 9} = \frac{(x - 3)(x^2 + 3x + 9)}{x^2 + 3x + 9} = x - 3

Applying

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

where

a = x

and

b = 3

allows cancellation of the quadratic factor.

Multiply the simplified expressions

(x^2 + 9)(x - 3) = x^3 - 3x^2 + 9x - 27

Distribute each term of the binomials to get the expanded polynomial.

Expand the subtracted term in the numerator

x(x - 3)^2 = x(x^2 - 6x + 9) = x^3 - 6x^2 + 9x

Expand

(x - 3)^2 = x^2 - 6x + 9

and distribute

x

Subtract the two expanded expressions to simplify the entire numerator

(x^3 - 3x^2 + 9x - 27) - (x^3 - 6x^2 + 9x) = 3x^2 - 27 = 3(x^2 - 9)

Combine like terms;

x^3

and

9x

terms cancel out, leaving

3x^2 - 27

Divide the simplified numerator by the main denominator

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

Cancel the common factor

(x^2 - 9)

from numerator and denominator.

Simplifying complex algebraic expressions via polynomial factoring (difference of squares and difference of cubes) and combining like terms.

Question