Question

The set of all real values of \(x\) for which the expansion of: $$ \left( 125x^2 - \frac{27}{x} \right)^{-23} $$ is valid, is: 

• A. \(\left(-\frac{3}{5}, \frac{3}{5}\right)\)
• B. \(\left(-\infty, -\frac{3}{5}\right) \cup \left(\frac{3}{5}, \infty\right)\)
• C. \(\left(-\frac{5}{3}, \frac{5}{3}\right)\)
• D. \(\left(-1, -\frac{1}{3}\right) \cup \left(\frac{1}{3}, 1\right)\)

Hint:

Ensure to account for all restrictions and solve inequalities carefully when dealing with radical expressions to correctly determine the domain of valid real values.

Answer 1

The Correct Option is B

The expression \(\left( 125x^2 - \frac{27}{x} \right)^{-\frac{2}{3}}\) is defined and real for all \(x\) where \(125x^2 - \frac{27}{x}\) is non-negative. 

Step 1: Simplify the expression and set up the inequality. First, simplify the expression: \[ 125x^2 - \frac{27}{x} = 125x^2 - 27x^{-1} \] To ensure the base of the power is non-negative: \[ 125x^2 - 27x^{-1} \geq 0 \] 

Step 2: Solve the inequality. To find the values of \(x\) satisfying this inequality: \[ 125x^3 - 27 \geq 0 \] Solving this inequality: \[ x^3 \geq \frac{27}{125} \] \[ x \geq \sqrt[3]{\frac{27}{125}} \text{ or } x \leq -\sqrt[3]{\frac{27}{125}} \] Calculating the cube root: \[ x \geq \frac{3}{5} \text{ or } x \leq -\frac{3}{5} \]

 Step 3: Refine the domain. Since \(x\) cannot be zero, the domain of \(x\) that satisfies the inequality, ensuring the entire expression is valid (not just non-negative), is: \[ x \in \left(-\infty, -\frac{3}{5}\right) \cup \left(\frac{3}{5}, \infty\right) \]