For all real numbers x, the algebraic expression \frac{x^6 - 27}{x^4 + 3x^2 + 9} can be si… — GRE General Test

For all real numbers

x

, the algebraic expression

\frac{x^6 - 27}{x^4 + 3x^2 + 9}

can be simplified to the polynomial form

ax^2 + bx + c

, where

a

b

, and

c

are real constants. What is the value of

a + b + c

Answer: -2

The simplified expression is

x^2 - 3

, which corresponds to polynomial coefficients

a = 1

b = 0

, and

c = -3

. The sum

a + b + c

equals

-2

Factoring the numerator

x^6 - 27

as a difference of cubes

(x^2)^3 - 3^3

produces

(x^2 - 3)(x^4 + 3x^2 + 9)

. Canceling the non-zero factor

(x^4 + 3x^2 + 9)

from the numerator and denominator simplifies the expression to

x^2 - 3

. In standard form

ax^2 + bx + c

a = 1

b = 0

, and

c = -3

. Adding these coefficients gives

1 + 0 + (-3) = -2

Factor the numerator using the difference of cubes formula

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

The expression

x^6 - 27

matches the pattern

u^3 - v^3

with

u = x^2

and

v = 3

Simplify the rational expression by canceling the common quadratic-biquadratic factor

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

Since

x^4 + 3x^2 + 9 > 0

for all real numbers

x

, the denominator is never zero, allowing direct cancellation of the common factor.

Match coefficients with

ax^2 + bx + c

and calculate

a + b + c

a = 1, b = 0, c = -3 \implies a + b + c = 1 + 0 + (-3) = -2

Comparing

x^2 - 3 = 1x^2 + 0x - 3

ax^2 + bx + c

determines the values of constants

a

b

, and

c

Difference of Cubes Factoring Identity

Estimated Time:1m 30s

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