We are given the equation:

$\sqrt[4]{(x^2 - x)^3} = 8$

Step 1: Eliminate the fourth root

To eliminate the fourth root, raise both sides of the equation to the power of 4:

$\left(\sqrt[4]{(x^2 - x)^3}\right)^4 = 8^4$

This simplifies to:

$(x^2 - x)^3 = 8^4$

Step 2: Calculate $8^4$

Since $8^4 = (2^3)^4 = 2^{12} = 4096$, we have:

$(x^2 - x)^3 = 4096$

Step 3: Take the cube root of both sides

To isolate $x^2 - x$, take the cube root of both sides:

$x^2 - x = \sqrt[3]{4096}$

Since $4096 = 16^3$, we find:

$x^2 - x = 16$

Step 4: Solve the quadratic equation

Now, solve the quadratic equation $x^2 - x = 16$:

$x^2 - x - 16 = 0$

Use the quadratic formula to solve for $x$:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-16)}}{2(1)}$ $x = \frac{1 \pm \sqrt{1 + 64}}{2}$ $x = \frac{1 \pm \sqrt{65}}{2}$

Thus, the two solutions for $x$ are:

$x = \frac{1 + \sqrt{65}}{2} \quad \text{or} \quad x = \frac{1 - \sqrt{65}}{2}$

Final Answer:

The two solutions to the equation are:

$x = \frac{1 + \sqrt{65}}{2} \quad \text{or} \quad x = \frac{1 - \sqrt{65}}{2}$

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Would you like a more detailed explanation on any of the steps? Here are some related questions for further practice:

1. What is the process for solving cubic and quartic equations?
2. How do you handle roots and exponents together in equations?
3. How can we graph the function $y = \sqrt[4]{(x^2 - x)^3}$?
4. How do we check if both solutions to the quadratic are valid in the original equation?
5. Can the equation $\sqrt[4]{(x^2 - x)^3} = 8$ have complex solutions?

Tip: Always check solutions in the original equation to ensure that they do not lead to extraneous results!