- The equation is solved by isolating the radical: \((4x)^{1/3} = x\).
- Both sides are cubed to remove the radical: \(4x = x^3\), resulting in \(x^3 - 4x = 0\).
- Factoring yields \(x(x-2)(x+2) = 0\), identifying the potential roots \(x = 0, 2, -2\).
- Verification in the original equation confirms all roots are valid, so the solutions are \(\boxed{{x \in \{-2, 0, 2\}}}\).
Explanation
- Analyzing the Equation and Initial Step
The objective is to solve the equation \((4x)^{1/3} - x = 0\). This equation contains a cube root term and a linear term. To isolate the variables and solve for \(x\), we need to follow a structured algebraic approach. First, we identify the structure of the equation: it is a radical equation where we can isolate the radical term. The equation is \((4x)^{1/3} = x\). By isolating the cube root, we prepare the equation to be solved by removing the exponent of \(1/3\).
- Algebraic Manipulation and Reducing to a Polynomial
Once the term \((4x)^{1/3}\) is isolated on one side, we can eliminate the cube root by cubing both sides of the equation. Performing this operation gives:
\[((4x)^{1/3})^3 = x^3\]
\[4x = x^3\]
This step transforms the radical equation into a polynomial equation, which is easier to work with. Rearranging terms leads to:
\[x^3 - 4x = 0\]
This is a standard cubic polynomial equation.
- Solving the Polynomial Equation
Now, we factor the polynomial \(x^3 - 4x = 0\). We can factor out \(x\):
\[x(x^2 - 4) = 0\]
Using the difference of squares formula, \(x^2 - 4 = (x-2)(x+2)\), we can further factor the equation:
\[x(x-2)(x+2) = 0\]
Setting each factor to zero, we find the potential roots: \(x = 0\), \(x = 2\), and \(x = -2\).
- Checking the Solutions and Conclusion
Finally, we check these potential solutions in the original equation \((4x)^{1/3} - x = 0\):
- For \(x = 0\): \((4 \times 0)^{1/3} - 0 = 0 - 0 = 0\). This is valid.
- For \(x = 2\): \((4 \times 2)^{1/3} - 2 = \sqrt[3]{8} - 2 = 2 - 2 = 0\). This is valid.
- For \(x = -2\): \((4 \times (-2))^{1/3} - (-2) = \sqrt[3]{-8} + 2 = -2 + 2 = 0\). This is valid.
All three potential solutions satisfy the original equation. Thus, the solutions are \(x \in \{-2, 0, 2\}\).
Examples
This problem demonstrates the importance of algebraic manipulation when dealing with radicals. In real-world engineering or physics, such equations often arise when calculating equilibrium positions or critical points where rates of change balance out. For example, if you were modeling the displacement of a structural component under varying stress, you might arrive at an equation like this to find the points where the total force is zero. Learning these steps—isolating, raising to powers, and verifying roots—provides a robust method for solving similar nonlinear systems.