Solve the equation. \( (6 x-1)^{2 / 3}=x^{1 / 3} \) The solution set is \( \{ \) (Simplify your answer. Type an integer complex numbers in terms of \( i \) )

To solve the equation \( (6x - 1)^{2/3} = x^{1/3} \), we can start by raising both sides to the power of \( 3 \) to eliminate the fractional exponents: \[ (6x - 1)^{2} = x \] Expanding the left side, we have: \[ (6x - 1)(6x - 1) = x \] This simplifies to: \[ 36x^2 - 12x + 1 = x \] Rearranging gives: \[ 36x^2 - 13x + 1 = 0 \] Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 36 \), \( b = -13 \), and \( c = 1 \): Calculating the discriminant: \[ b^2 - 4ac = (-13)^2 - 4(36)(1) = 169 - 144 = 25 \] Then we find the roots: \[ x = \frac{13 \pm \sqrt{25}}{2 \cdot 36} = \frac{13 \pm 5}{72} \] This gives us two solutions: 1. \( x = \frac{18}{72} = \frac{1}{4} \) 2. \( x = \frac{8}{72} = \frac{1}{9} \) Next, we need to verify if these solutions satisfy the original equation, especially since raising to an even power can introduce extraneous solutions. For \( x = \frac{1}{4} \): \[ (6(\frac{1}{4}) - 1)^{2/3} = (1.5)^{2/3} \quad \text{and} \quad (\frac{1}{4})^{1/3} = (0.25)^{1/3} \] Evaluate \( 6(\frac{1}{4}) - 1 = 1.5 \): \[ (1.5)^{2/3} = \sqrt[3]{(1.5)^2} = \sqrt[3]{2.25} \] For \( x = \frac{1}{9} \): \[ (6(\frac{1}{9}) - 1)^{2/3} = (-\frac{1}{3})^{2/3} = \frac{1}{\sqrt[3]{9}} \quad \text{and} \quad (\frac{1}{9})^{1/3} = \frac{1}{3} \] Substituting back validates that \( x = \frac{1}{4} \) fits while \( x = \frac{1}{9} \) does not yield a valid output due to extraneous roots. Thus, the solution set is: \{ \(\frac{1}{4}\) \}