We are tasked with solving the equation:

$8x^2 (x^2 - 4)^{1/3} + (x^2 - 4)^{4/3} = 0$

Step-by-Step Solution:

Step 1: Factorize the equation

The common factor in both terms is $(x^2 - 4)^{1/3}$. Factoring it out gives: $(x^2 - 4)^{1/3} \left[ 8x^2 + (x^2 - 4) \right] = 0$

Step 2: Analyze the factors

The equation will equal zero if either factor equals zero:

1. $(x^2 - 4)^{1/3} = 0$
2. $8x^2 + (x^2 - 4) = 0$

---

Case 1: $(x^2 - 4)^{1/3} = 0$

For this to hold, $x^2 - 4 = 0$, which simplifies to: $x^2 = 4 \implies x = \pm 2$

---

Case 2: $8x^2 + (x^2 - 4) = 0$

Simplify the expression inside the brackets: $8x^2 + x^2 - 4 = 0 \implies 9x^2 - 4 = 0$ Factoring $9x^2 - 4$ as a difference of squares: $(3x - 2)(3x + 2) = 0$ This gives two solutions: $x = \frac{2}{3} \quad \text{and} \quad x = -\frac{2}{3}$

---

Final Solution:

Combining the results from both cases, the solutions are: $x = \pm 2, \quad x = \frac{2}{3}, \quad \text{and} \quad x = -\frac{2}{3}$

---

Verification:

You can verify by substituting each solution into the original equation.

---

Would you like a detailed explanation of the verification or clarification on any step?

Related Questions:

1. How can you simplify equations involving fractional exponents systematically?
2. What are the steps for factoring polynomials with higher powers?
3. How do you verify solutions of equations by substitution?
4. What are common methods to handle equations involving radicals?
5. Can all equations with fractional exponents be converted to polynomial form?

Tip:

When dealing with fractional exponents, isolating the base with the exponent first often simplifies the process.