<Question>
Solve for the positive value of $x$ such that $\sqrt[3]{x^2 - 4x + 4} = 16$.
</Question>

<think>
<Plan_1>
Okay, so I need to solve for the positive value of x in the equation where the cube root of (x squared minus 4x plus 4) equals 16. Let me write that down again to make sure I have it right: ∛(x² - 4x + 4) = 16. Alright, the goal is to find the positive x that satisfies this equation.

First, since there's a cube root on the left side, maybe I should get rid of that by cubing both sides of the equation. That way, I can eliminate the cube root and simplify things. Let me try that.
So if I cube both sides, the left side becomes (∛(x² - 4x + 4))³, which simplifies to just x² - 4x + 4. And the right side would be 16 cubed.

Let me compute 16 cubed.

<prompt_1.1> Compute 16 cubed and write the transformed equation after cubing both sides. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
16 squared is 256, so 16 cubed is 256 multiplied by 16. Let me do that multiplication. 256 times 10 is 2560, and 256 times 6 is 1536. Adding those together gives 2560 + 1536 = 4096. So, 16 cubed is 4096. Therefore, the equation becomes x² - 4x + 4 = 4096.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Now, I need to solve this quadratic equation. Let me rearrange it so that one side is zero.

<prompt_2.1> Rearrange the equation to form a standard quadratic equation. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Subtracting 4096 from both sides gives x² - 4x + 4 - 4096 = 0. Simplifying the constants: 4 - 4096 is -4092. So the equation is x² - 4x - 4092 = 0.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Hmm, quadratic equations. The standard form is ax² + bx + c = 0. Here, a is 1, b is -4, and c is -4092. To solve for x, I can use the quadratic formula: x = [-b ± √(b² - 4ac)] / (2a). Let me plug in the values.

<prompt_3.1> Solve the quadratic equation using the quadratic formula. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
First, calculate the discriminant, which is b² - 4ac. b squared is (-4)² = 16. Then, 4ac is 4 * 1 * (-4092) = -16368. So the discriminant becomes 16 - (-16368) = 16 + 16368 = 16384. Wait, 16384? Let me check that again. If it's b² - 4ac, and a is 1, c is -4092, so 4ac is 4*1*(-4092) = -16368. So subtracting that is like adding 16368. So 16 + 16368 is indeed 16384. Hmm, 16384. That seems like a perfect square. Let me confirm.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Let's see, 128 squared is 16384 because 100 squared is 10000, 80 squared is 6400, so 80 squared is 6400, 120 squared is 14400, 130 squared is 16900. So somewhere between 120 and 130. Let's try 128: 128*128.

<prompt_4.1> Confirm that 128 squared equals 16384 and use it to find the square root of the discriminant. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
128*100=12800, 128*28=3584, so 12800+3584=16384. Yes, that's right. So the square root of 16384 is 128. Therefore, the square root part is 128.

Now plug back into the quadratic formula: x = [4 ± 128] / 2. Because -b is -(-4) which is 4. So the two solutions are (4 + 128)/2 and (4 - 128)/2. Let's compute both.

First solution: (4 + 128) = 132. 132 divided by 2 is 66. Second solution: (4 - 128) = -124. -124 divided by 2 is -62. So the solutions are x = 66 and x = -62.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But the problem specifically asks for the positive value of x. So between 66 and -62, the positive one is 66. Therefore, x = 66.

But hold on, I need to check if this solution actually works in the original equation. Sometimes when we square both sides or cube both sides, we can introduce extraneous solutions.

<prompt_5.1> Verify the solution by substituting x = 66 back into the original equation. </prompt_5.1>

But wait, just to be thorough, let's check the other solution as well, even though the problem asks for the positive one. If we plug x = -62 into the original equation: ∛((-62)^2 - 4*(-62) + 4). Let's compute inside the cube root.

<prompt_5.2> Verify the solution by substituting x = -62 back into the original equation. </prompt_5.2>
-----

Wait, but let me double-check my steps because sometimes even if you cube both sides, maybe there was an error in expanding the quadratic. Let me verify each step again.

<prompt_5.3> Re-evaluate the steps to ensure no errors were made in expanding or solving the quadratic equation. </prompt_5.3>

But let me just think again: is there another way this equation could be approached? Maybe factoring the quadratic? Let's see, x² -4x -4092 =0.

<prompt_5.4> Explore alternative methods to solve the equation and confirm the solution. </prompt_5.4>
</Plan_5>

<Execution_5>
<execution_5.1>
Original equation: ∛(x² -4x +4) =16. Let's substitute x=66. Compute the expression inside the cube root: 66 squared is 4356, minus 4*66 which is 264, plus 4. So 4356 - 264 +4. Let's compute 4356 -264: 4356 - 200 is 4156, then -64 more is 4092. Then add 4: 4092 +4 = 4096. So the cube root of 4096 is indeed 16, because 16 cubed is 4096. So that checks out.
</execution_5.1>

<execution_5.2>
First, (-62)^2 is 3844. Then, -4*(-62) is +248. Adding 4. So total inside the cube root is 3844 + 248 + 4. Let's compute that. 3844 + 248 is 4092, plus 4 is 4096. So cube root of 4096. But 4096 is 16 cubed, right? Wait, 16^3 is 4096. So cube root of 4096 is 16. So the left side is 16, which equals the right side. So x = -62 is also a valid solution.
</execution_5.2>

<execution_5.3>
Original equation: ∛(x² -4x +4) =16. Cube both sides: x² -4x +4 =16³=4096. Then x² -4x +4 -4096=0 => x² -4x -4092=0. Discriminant: (-4)^2 -4*1*(-4092)=16 +16368=16384. sqrt(16384)=128. Solutions: [4 ±128]/2. So 132/2=66 and -124/2=-62.
</execution_5.3>

<execution_5.4>
Factoring a quadratic with such a large constant term might be tricky, but perhaps. Let me see if there are two numbers that multiply to -4092 and add up to -4. Hmm. The factors of 4092. Let's see, 4092 divided by 4 is 1023. 1023 is divisible by 3: 1023/3=341. 341 is 11*31. So prime factors of 4092 are 2*2*3*11*31. So possible pairs. Let's see, we need two numbers whose product is -4092 and sum is -4. So one is positive, one is negative. Let me think. Let's try 62 and 66. Wait, 62*66=4092. So if we have 66 and -62, their product is -4092, and their sum is 4. But we need sum -4. So maybe -66 and 62? Then sum is -4. Yes! -66 +62= -4, and (-66)*(62)= -4092. Perfect. So the quadratic factors as (x - 66)(x + 62) =0. Therefore, x=66 or x=-62. So that's the same result as before. So that confirms the solutions. So 66 is the positive one. So that's correct.

Therefore, the answer is 66. But just to make sure, let's plug x=66 into the original equation. Compute x² -4x +4. That's 66² -4*66 +4. 66 squared is 4356. 4*66 is 264. So 4356 -264 +4. 4356 -264 is 4092, plus 4 is 4096. Then the cube root of 4096 is 16, since 16 cubed is 4096. So that works. Perfect. So x=66 is indeed the solution.

So, after going through the steps, checking the discriminant, solving the quadratic, factoring it, and verifying the solution in the original equation, I can confidently say that the positive value of x is 66.
</execution_5.4>
</Execution_5>

<Plan_6>
Based on execution_5.1:
But wait, let me check if there's any mistake in my steps. Let's go through the process again.

<prompt_6.1> Re-evaluate the steps to ensure the solution is correct. </prompt_6.1>
-----

Based on execution_5.4:
**Final Answer**
The positive solution is \boxed{66}.
</Plan_6>

<Execution_6>
<execution_6.1>
Starting with the original equation: cube root of (x² -4x +4) =16. Cube both sides: x² -4x +4 =16³. 16 cubed is 4096. Then x² -4x +4 -4096=0 simplifies to x² -4x -4092=0. Quadratic formula: [4 ± sqrt(16 +16368)] /2. sqrt(16384)=128. So [4 ±128]/2. That gives 132/2=66 and -124/2=-62. Check x=66: inside cube root is 66² -4*66 +4. 66² is 4356, 4*66 is 264, so 4356 -264 +4=4096. Cube root of 4096 is 16. Correct. Check x=-62: (-62)^2=3844, -4*(-62)=248, so 3844 +248 +4=4096. Cube root is 16. So both solutions are valid.
</execution_6.1>
</Execution_6>

<Final_answer>
Therefore, the positive solution is $\boxed{66}$.
</Final_answer>
</think>