Question

select the correct answer from each drop-down menu. consider this equation. ((4x)^{\frac{1}{3}} - x = 0) the first step in solving this equation is to add x to both sides. the second step is to cube both sides. solving this equation for x initially yields three possible values, one possible value, two possible values. checking the solutions shows that

Explanation:

Step1: First step analysis

The equation is \((4x)^{\frac{1}{3}} - x = 0\). To isolate the radical term, we add \(x\) to both sides. So the first step is to add \(x\) to both sides.

Step2: Second step analysis

After adding \(x\) to both sides, we get \((4x)^{\frac{1}{3}} = x\). To eliminate the cube - root, we cube both sides. So the second step is to cube both sides.

Step3: Solve the equation after cubing

Cubing both sides of \((4x)^{\frac{1}{3}} = x\) gives \(4x=x^{3}\). Rearranging this equation, we have \(x^{3}-4x = 0\). Factoring out an \(x\), we get \(x(x^{2}-4)=0\). Then, using the difference of squares \(a^{2}-b^{2}=(a + b)(a - b)\), we can factor \(x^{2}-4\) as \((x + 2)(x - 2)\). So the equation becomes \(x(x + 2)(x - 2)=0\). The solutions are \(x = 0\), \(x=-2\), and \(x = 2\). But we need to check these solutions in the original equation.
- For \(x = 0\): \((4\times0)^{\frac{1}{3}}-0=0 - 0=0\), which is valid.
- For \(x=-2\): \((4\times(-2))^{\frac{1}{3}}-(-2)=\sqrt[3]{-8}+2=-2 + 2=0\), which is valid.
- For \(x = 2\): \((4\times2)^{\frac{1}{3}}-2=\sqrt[3]{8}-2=2 - 2=0\), which is valid. Wait, but initially when we solve \(x^{3}-4x = 0\), we get three solutions, but when we check, all three are valid? Wait, no, let's re - check the original equation \((4x)^{\frac{1}{3}}-x = 0\) or \((4x)^{\frac{1}{3}}=x\). The domain of the cube - root function is all real numbers, but we need to ensure that the left - hand side and right - hand side are equal.
Wait, when \(x=-2\): \((4\times(-2))^{\frac{1}{3}}=\sqrt[3]{-8}=-2\), and the right - hand side is \(x=-2\), so it works. When \(x = 2\): \((4\times2)^{\frac{1}{3}}=\sqrt[3]{8}=2\), and the right - hand side is \(x = 2\), so it works. When \(x = 0\): \((4\times0)^{\frac{1}{3}}=0\), and the right - hand side is \(x = 0\), so it works. But wait, initially, when we solve the equation \(x^{3}-4x = 0\), we get three solutions. But let's go back to the step of cubing. Cubing is a non - one - to - one operation, but in this case, all the solutions we get from cubing satisfy the original equation. However, if we made a mistake in the first two steps, but according to the process:
After the first step (add \(x\)) and the second step (cube both sides), we solve the equation and get three possible values initially (before checking, when we solve \(x^{3}-4x = 0\), we have three roots), but when we check, all three are valid. Wait, but maybe I made a mistake. Wait, the original equation is \((4x)^{\frac{1}{3}}-x = 0\). Let's re - express the first step:
Original equation: \((4x)^{\frac{1}{3}}-x = 0\)
Step 1: Add \(x\) to both sides: \((4x)^{\frac{1}{3}}=x\)
Step 2: Cube both sides: \(4x=x^{3}\)
\(x^{3}-4x = 0\)
\(x(x^{2}-4)=0\)
\(x(x - 2)(x + 2)=0\)
So \(x = 0\), \(x = 2\), \(x=-2\)
Check \(x=-2\): \((4\times(-2))^{\frac{1}{3}}=\sqrt[3]{-8}=-2\), and \(-2-(-2)=0\), correct.
\(x = 2\): \(\sqrt[3]{8}-2=2 - 2=0\), correct.
\(x = 0\): \(\sqrt[3]{0}-0=0\), correct. So initially, when we solve the equation \(x^{3}-4x = 0\), we get three possible values, and when we check, all three are valid. But the problem says "initially yields" (before checking? No, when we solve the equation after the first two steps, we get three solutions, and when we check, all three are valid. Wait, maybe the question has a typo, but according to the process:
The first step: add \(x\) to both sides.
The second step: cube both sides.
When we solve the equation after these two steps, we initially (before checking) get three possible values, and when we check, all three are valid (in this case). But let's follow the problem's drop - down options. The first drop - down for the first step: "add x to both sides" is correct. The second drop - down for the second step: "cube both sides" is correct. The third drop - down: "three possible values" (since we get \(x = 0\), \(x=-2\), \(x = 2\) from solving \(x^{3}-4x = 0\) and all are valid when checked).

Answer:

First step: add x to both sides; Second step: cube both sides; Initially yields three possible values.