How would I solve this equation? [(x^(2/3)) - (5x^(1/3)) + 6 = 0]

To solve the equation \( x^{\frac{2}{3}} - 5x^{\frac{1}{3}} + 6 = 0 \), we can simplify the expression by making a substitution to turn it into a quadratic form. Here's how to approach the solution:

### Step 1: Substitution
Let:
\[
y = x^{\frac{1}{3}}
\]
This implies:
\[
y^2 = x^{\frac{2}{3}}
\]
Now, substitute \( y \) and \( y^2 \) into the original equation:
\[
y^2 - 5y + 6 = 0
\]

### Step 2: Solve the Quadratic Equation
We now have a quadratic equation in \( y \):
\[
y^2 - 5y + 6 = 0
\]
This can be factored as:
\[
(y - 2)(y - 3) = 0
\]

### Step 3: Solve for \( y \)
Setting each factor equal to 0:
1. \( y - 2 = 0 \) gives \( y = 2 \)
2. \( y - 3 = 0 \) gives \( y = 3 \)

### Step 4: Back-substitute for \( x \)
Recall that \( y = x^{\frac{1}{3}} \), so:
1. If \( y = 2 \), then \( x^{\frac{1}{3}} = 2 \), which implies \( x = 2^3 = 8 \).
2. If \( y = 3 \), then \( x^{\frac{1}{3}} = 3 \), which implies \( x = 3^3 = 27 \).

### Final Solutions:
The solutions to the equation are:
\[
x = 8 \quad \text{or} \quad x = 27
\]

Would you like further details on any of the steps?

### Related Questions:
1. How do you factor quadratic equations like \( y^2 - 5y + 6 = 0 \)?
2. How can you check if a solution to a cubic equation is correct?
3. What is the process for solving equations that contain fractional exponents?
4. How does substitution work for transforming non-quadratic equations into quadratic form?
5. What is the general method for solving polynomial equations of higher degrees?

### Tip:
When solving equations with exponents, substitution is a powerful technique to reduce the complexity and make factoring easier.

Discover how to solve the equation x^(2/3) - 5x^(1/3) + 6 = 0 by using substitution and factoring quadratic equations. This detailed guide covers algebraic concepts and methods involving fractional exponents and substitution, making it ideal for students in Grades 9-10.

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