To solve the equation:

$x^{\frac{2}{3}} - 5x^{\frac{1}{2}} + 6 = 0$

we can simplify the expression by making a substitution for clarity. Let's proceed step by step:

Step 1: Substitution

Define a new variable: $y = x^{\frac{1}{6}}$ This implies: $x^{\frac{2}{3}} = y^4 \quad \text{and} \quad x^{\frac{1}{2}} = y^3$

Thus, the original equation becomes: $y^4 - 5y^3 + 6 = 0$

Step 2: Solve the polynomial equation

We now solve the polynomial equation: $y^4 - 5y^3 + 6 = 0$ This is a quartic equation in terms of $y$.

Let's try factoring the polynomial: $y^4 - 5y^3 + 6 = (y^2 - 2)(y^2 - 3) = 0$

Step 3: Solve for $y$

Setting each factor to zero:

$y^2 - 2 = 0 \quad \Rightarrow \quad y^2 = 2 \quad \Rightarrow \quad y = \pm \sqrt{2}$ $y^2 - 3 = 0 \quad \Rightarrow \quad y^2 = 3 \quad \Rightarrow \quad y = \pm \sqrt{3}$

Step 4: Back-substitute $y = x^{\frac{1}{6}}$

Now, we substitute back $y = x^{\frac{1}{6}}$ into each solution:

1. If $y = \sqrt{2}$, then: $x^{\frac{1}{6}} = \sqrt{2} \quad \Rightarrow \quad x = (\sqrt{2})^6 = 2^3 = 8$

2. If $y = -\sqrt{2}$, then: $x^{\frac{1}{6}} = -\sqrt{2} \quad \Rightarrow \quad x = (-\sqrt{2})^6 = (-1)^6 \cdot (\sqrt{2})^6 = 8$

3. If $y = \sqrt{3}$, then: $x^{\frac{1}{6}} = \sqrt{3} \quad \Rightarrow \quad x = (\sqrt{3})^6 = 3^3 = 27$

4. If $y = -\sqrt{3}$, then: $x^{\frac{1}{6}} = -\sqrt{3} \quad \Rightarrow \quad x = (-\sqrt{3})^6 = 27$

Step 5: Final solutions

The solutions to the equation are:

$x = 8 \quad \text{and} \quad x = 27$

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Would you like more details, or do you have any questions about this method?

Related Questions:

1. How do we factor quartic equations like the one in Step 2?
2. What other substitution techniques could be used to solve similar equations?
3. How does the process change if negative exponents are involved?
4. What are the implications of multiple solutions in polynomial equations?
5. Can you graph this equation and show where the roots lie?

Tip: Always verify by plugging your solutions back into the original equation!