If 3^x + 3^{-x} = 4, what is the value of \frac{27^x + 27^{-x} - 2}{9^x + 9^{-x} + 1}? — GRE General Test

If $3^x + 3^{-x} = 4$ , what is the value of $\frac{27^x + 27^{-x} - 2}{9^x + 9^{-x} + 1}$ ?

\frac{10}{3}

Squaring

3^x + 3^{-x} = 4

yields

9^x + 2 + 9^{-x} = 16

, which simplifies to

9^x + 9^{-x} = 14

. Utilizing the sum of cubes identity gives

27^x + 27^{-x} = (3^x + 3^{-x})(9^x - 1 + 9^{-x}) = 4(14 - 1) = 52

. Substituting these values into the given fraction gives

\frac{52 - 2}{14 + 1} = \frac{50}{15} = \frac{10}{3}

Square the given expression

3^x + 3^{-x} = 4

to determine

9^x + 9^{-x}

(3^x + 3^{-x})^2 = 9^x + 2(3^x)(3^{-x}) + 9^{-x} = 9^x + 2 + 9^{-x} = 16

, which yields

9^x + 9^{-x} = 14

Expanding the square of a binomial requires accounting for the middle term

2 \cdot 3^x \cdot 3^{-x} = 2

Express

27^x + 27^{-x}

using the sum of cubes factorization identity.

27^x + 27^{-x} = (3^x)^3 + (3^{-x})^3 = (3^x + 3^{-x})(9^x - 3^x \cdot 3^{-x} + 9^{-x}) = 4(14 - 1) = 52

The identity

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

allows factoring cubic exponential expressions.

Substitute the evaluated terms into the target rational expression and simplify.

\frac{27^x + 27^{-x} - 2}{9^x + 9^{-x} + 1} = \frac{52 - 2}{14 + 1} = \frac{50}{15} = \frac{10}{3}.

Replacing component expressions with their computed values yields the simplified numerical fraction.

Evaluating high-power exponential expressions using polynomial identity transformations and exponent laws.

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