For all real numbers x \neq -5, the algebraic expression \frac{x^3 + 125}{x^2 - 5x + 25} -… — GRE General Test

For all real numbers

x \neq -5

, the algebraic expression

\frac{x^3 + 125}{x^2 - 5x + 25} - \frac{x^2 - 25}{x + 5}

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

Answer: 10

The simplified value of the expression for all valid real numbers x is 10.

Factoring the numerators reveals that the first term reduces to

x + 5

and the second term reduces to

x - 5

. Subtracting

(x - 5)

from

(x + 5)

yields

(x + 5) - (x - 5) = 10

, which is constant for all valid values of

x

Factor the numerator of the first rational term using the sum of cubes identity.

x^3 + 125 = (x + 5)(x^2 - 5x + 25)

The sum of cubes formula

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

applies with

a = x

and

b = 5

Cancel the non-zero factor

(x^2 - 5x + 25)

from the first fraction.

\frac{(x + 5)(x^2 - 5x + 25)}{x^2 - 5x + 25} = x + 5

The quadratic factor

x^2 - 5x + 25

has a negative discriminant (

(-5)^2 - 4(1)(25) = -75 < 0

), so it is never zero for any real number

x

Factor the numerator of the second rational term using the difference of squares identity.

x^2 - 25 = (x - 5)(x + 5)

The difference of squares formula

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

applies with

a = x

and

b = 5

Cancel the common factor

(x + 5)

from the second fraction.

\frac{(x - 5)(x + 5)}{x + 5} = x - 5

Given

x \neq -5

, the factor

x + 5

is non-zero and can be canceled.

Subtract the two simplified terms.

(x + 5) - (x - 5) = x + 5 - x + 5 = 10

Distribute the negative sign to both terms in

(x - 5)

and combine like terms.

Factoring sum of cubes and difference of squares to simplify rational expressions

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