We are given the equation:

$\left( x - 2 \right)^{\frac{2}{3}} + \left( x - 2 \right)^{\frac{1}{3}} - 12 = 0$

To solve this, let's introduce a substitution to simplify the expression. Set:

$y = \left( x - 2 \right)^{\frac{1}{3}}$

This substitution allows us to rewrite the original equation as:

$y^2 + y - 12 = 0$

Now, we can solve this quadratic equation for $y$ using the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For $y^2 + y - 12 = 0$, the coefficients are $a = 1$, $b = 1$, and $c = -12$. Substituting into the quadratic formula:

$y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-12)}}{2(1)}$ $y = \frac{-1 \pm \sqrt{1 + 48}}{2}$ $y = \frac{-1 \pm \sqrt{49}}{2}$ $y = \frac{-1 \pm 7}{2}$

This gives two solutions for $y$:

$y = \frac{-1 + 7}{2} = 3 \quad \text{or} \quad y = \frac{-1 - 7}{2} = -4$

Now, recall that $y = \left( x - 2 \right)^{\frac{1}{3}}$, so we can solve for $x$ by cubing both sides.

Case 1: $y = 3$

$\left( x - 2 \right)^{\frac{1}{3}} = 3$ Cubing both sides: $x - 2 = 27$ $x = 27 + 2 = 29$

Case 2: $y = -4$

$\left( x - 2 \right)^{\frac{1}{3}} = -4$ Cubing both sides: $x - 2 = (-4)^3 = -64$ $x = -64 + 2 = -62$

Thus, the solutions to the equation are: $x = 29 \quad \text{or} \quad x = -62$

Would you like a detailed explanation of any specific step, or have any questions?

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Here are five questions related to the problem:

1. How does substitution simplify complex equations involving fractional exponents?
2. What are the advantages of using the quadratic formula to solve quadratic equations?
3. Why do we cube both sides when reversing the substitution for $y$?
4. What are the conditions under which the cubic root function is valid for negative numbers?
5. How can you verify the solutions to this type of equation?

Tip: When solving fractional exponent equations, always check if the final solutions satisfy the original equation, especially with negative or zero exponents.