How do you solve [(x^(7/3)) - (3x^(4/3)) - (4x^(1/3)) = 0] ?

To solve the equation \((x^{7/3}) - 3x^{4/3} - 4x^{1/3} = 0\), we can approach it by simplifying and factoring. Here's a step-by-step solution:

### Step 1: Substitution
Let’s introduce a substitution to simplify the powers of \(x\). Define:
\[
y = x^{1/3}
\]
This implies that:
- \(x^{7/3} = y^7\)
- \(x^{4/3} = y^4\)
- \(x^{1/3} = y\)

Now, the equation becomes:
\[
y^7 - 3y^4 - 4y = 0
\]

### Step 2: Factor the equation
We can now factor the equation. Notice that \(y\) is a common factor in all the terms:
\[
y(y^6 - 3y^3 - 4) = 0
\]

This gives us two cases to consider:

### Case 1: \(y = 0\)
If \(y = 0\), then:
\[
x^{1/3} = 0 \quad \Rightarrow \quad x = 0
\]

### Case 2: \(y^6 - 3y^3 - 4 = 0\)
Let’s make another substitution to simplify further. Let:
\[
z = y^3
\]
This turns the equation into:
\[
z^2 - 3z - 4 = 0
\]

### Step 3: Solve the quadratic equation
Now, solve the quadratic equation:
\[
z^2 - 3z - 4 = 0
\]
Using the quadratic formula:
\[
z = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)}
\]
\[
z = \frac{3 \pm \sqrt{9 + 16}}{2}
\]
\[
z = \frac{3 \pm \sqrt{25}}{2}
\]
\[
z = \frac{3 \pm 5}{2}
\]
So, the two solutions are:
\[
z = \frac{3 + 5}{2} = 4 \quad \text{or} \quad z = \frac{3 - 5}{2} = -1
\]

### Step 4: Return to \(y\) and solve for \(x\)
Since \(z = y^3\), we have two cases:
1. \(y^3 = 4\), so \(y = \sqrt[3]{4}\).
2. \(y^3 = -1\), so \(y = \sqrt[3]{-1} = -1\).

Now, return to \(x = y^3\):
1. For \(y = \sqrt[3]{4}\), we get \(x = (\sqrt[3]{4})^3 = 4\).
2. For \(y = -1\), we get \(x = (-1)^3 = -1\).

### Final solutions:
Thus, the solutions to the equation are:
\[
x = 0, \, 4, \, -1
\]

Would you like further details on any part of the process? 

### Related questions:
1. How do you factor cubic equations like this one?
2. What is the general method for solving higher-degree polynomials?
3. How does the cubic root function behave for negative inputs?
4. How can the quadratic formula be derived from a general quadratic equation?
5. What are common factorization techniques for polynomials?

### Tip:
Substitution is a powerful method in algebra to reduce complex equations into simpler forms by introducing new variables.

How do you solve [(x^(7/3)) - (3x^(4/3)) - (4x^(1/3)) = 0]?

Discover how to solve the polynomial equation x^(7/3) - 3x^(4/3) - 4x^(1/3) = 0 using substitution, factoring, and the quadratic formula. This detailed solution explains key concepts in algebra, including higher-degree polynomials and factorization. Ideal for students in Grades 10-12.

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