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Solve the following equation by factoring

The equation provided is:

\[
(5x - 7)^{\frac{2}{3}} - 6(5x - 7)^{\frac{1}{3}} = 0
\]

### Steps to solve by factoring:

1. **Substitute** \( y = (5x - 7)^{\frac{1}{3}} \). This simplifies the equation into a quadratic form:
   \[
   y^2 - 6y = 0
   \]

2. **Factor** the quadratic equation:
   \[
   y(y - 6) = 0
   \]

3. **Solve for \( y \)**:
   \[
   y = 0 \quad \text{or} \quad y = 6
   \]

4. **Substitute back** \( y = (5x - 7)^{\frac{1}{3}} \) into each solution:
   - For \( y = 0 \):
     \[
     (5x - 7)^{\frac{1}{3}} = 0 \quad \Rightarrow \quad 5x - 7 = 0 \quad \Rightarrow \quad x = \frac{7}{5}
     \]
   - For \( y = 6 \):
     \[
     (5x - 7)^{\frac{1}{3}} = 6 \quad \Rightarrow \quad 5x - 7 = 6^3 = 216 \quad \Rightarrow \quad 5x = 223 \quad \Rightarrow \quad x = \frac{223}{5}
     \]

### Final Solutions:
\[
x = \frac{7}{5} \quad \text{or} \quad x = \frac{223}{5}
\]

Let me know if you want more details or have any questions!

Here are some related questions:
1. How do you recognize a cubic root equation in other problems?
2. What are the general steps for solving radical equations?
3. How do you verify solutions for equations involving fractional powers?
4. Can all equations with fractional exponents be solved by substitution?
5. What are other methods to solve quadratic-like equations?

**Tip**: Always check if solutions satisfy the original equation to avoid extraneous solutions caused by squaring.

Solve the following equation by factoring: (5x - 7)^(2/3) - 6(5x - 7)^(1/3) = 0

Learn how to solve the equation (5x - 7)^(2/3) - 6(5x - 7)^(1/3) = 0 by factoring. This problem involves key algebraic concepts, including substitution for radical equations and solving quadratic equations. Suitable for students in Grades 10-12.

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