Answer the following questions :

$\textbf{(a)}$  Find all real solutions of the equation $$\Big(x^2-2x\Big)^{x^2+x-6}=1$$ Explain why your solutions are the only solutions.

$\textbf{(b)}$  The following expression is a rational number. Find its value. $$\sqrt[3]{6\sqrt{3}+10} -\sqrt[3]{6\sqrt{3}-10}$$
### Part (a)

We need to find all real solutions of the equation:
\[
\left(x^2 - 2x\right)^{x^2 + x - 6} = 1
\]

To solve this, we consider the properties of exponents. The equation \(a^b = 1\) holds if one of the following conditions is true:
1. \(a = 1\)
2. \(a = -1\) and \(b\) is an even integer
3. \(b = 0\)

#### Case 1: \(x^2 - 2x = 1\)

Solving the quadratic equation:
\[
x^2 - 2x - 1 = 0
\]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[
x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}
\]
Thus, the solutions for this case are:
\[
x = 1 + \sqrt{2} \quad \text{and} \quad x = 1 - \sqrt{2}
\]

#### Case 2: \(x^2 + x - 6 = 0\)

Solving the quadratic equation:
\[
x^2 + x - 6 = 0
\]
Using the quadratic formula:
\[
x = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}
\]
Thus, the solutions for this case are:
\[
x = 2 \quad \text{and} \quad x = -3
\]

However, we need to check if these solutions satisfy the original equation. For \(x = 2\):
\[
(2^2 - 2 \cdot 2)^{2^2 + 2 - 6} = 0^0
\]
Since \(0^0\) is indeterminate, \(x = 2\) is not a valid solution.

For \(x = -3\):
\[
((-3)^2 - 2 \cdot (-3))^{(-3)^2 - 3 - 6} = (9 + 6)^0 = 15^0 = 1
\]
Thus, \(x = -3\) is a valid solution.

#### Case 3: \(x^2 - 2x = -1\)

Solving the quadratic equation:
\[
x^2 - 2x + 1 = 0 \implies (x - 1)^2 = 0 \implies x = 1
\]
Thus, the solution for this case is:
\[
x = 1
\]

Combining all valid solutions, we have:
\[
x = -3, \quad x = 1 + \sqrt{2}, \quad x = 1 - \sqrt{2}, \quad x = 1
\]

### Part (b)

We need to find the value of:
\[
\sqrt[3]{6\sqrt{3} + 10} - \sqrt[3]{6\sqrt{3} - 10}
\]

Let:
\[
a = \sqrt[3]{6\sqrt{3} + 10} \quad \text{and} \quad b = \sqrt[3]{6\sqrt{3} - 10}
\]

We need to find \(a - b\). Assume:
\[
a = \sqrt{3} + 1 \quad \text{and} \quad b = \sqrt{3} - 1
\]

Cubing both sides:
\[
a^3 = (6\sqrt{3} + 10) \quad \text{and} \quad b^3 = (6\sqrt{3} - 10)
\]

Then:
\[
a^3 - b^3 = (6\sqrt{3} + 10) - (6\sqrt{3} - 10) = 20
\]

Using the identity for the difference of cubes:
\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]

We need to find \(a^2 + ab + b^2\):
\[
a^2 = (\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}
\]
\[
b^2 = (\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}
\]
\[
ab = (\sqrt{3} + 1)(\sqrt{3} - 1) = 3 - 1 = 2
\]
\[
a^2 + ab + b^2 = (4 + 2\sqrt{3}) + 2 + (4 - 2\sqrt{3}) = 10
\]

Thus:
\[
20 = (a - b) \cdot 10 \implies a - b = 2
\]

The final answer is \(\boxed{2}\)