The equation is \( x^6 + x^3 - 2 = 0 \).

Let's solve this step by step.

1. Substitute \( y = x^3 \) to simplify the equation:
   \[
   y^2 + y - 2 = 0
   \]
   This is now a quadratic equation in terms of \( y \).

2. Solve the quadratic equation using the quadratic formula:
   \[
   y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = 1 \), \( b = 1 \), and \( c = -2 \).

   \[
   y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm \sqrt{9}}{2}
   \]
   \[
   y = \frac{-1 \pm 3}{2}
   \]

   So, \( y = 1 \) or \( y = -2 \).

3. Now, substitute back \( y = x^3 \):
   - If \( x^3 = 1 \), then \( x = 1 \).
   - If \( x^3 = -2 \), solve \( x = \sqrt[3]{-2} \). The solutions are:
     \[
     x = \sqrt[3]{-2}, \quad x = \sqrt[3]{-2} \cdot e^{i 2\pi/3}, \quad x = \sqrt[3]{-2} \cdot e^{i 4\pi/3}.
     \]
     These are the three cube roots of \( -2 \).

### Final Solutions:
1. \( x = 1 \)
2. \( x = \sqrt[3]{-2} \)
3. \( x = \sqrt[3]{-2} \cdot e^{i 2\pi/3} \)
4. \( x = \sqrt[3]{-2} \cdot e^{i 4\pi/3} \)

Would you like to go deeper into the complex roots or need a different explanation?

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Here are 5 related questions to expand:
1. How do you find cube roots of complex numbers?
2. Can we solve similar higher-degree polynomial equations?
3. What are the properties of complex numbers used in solving this?
4. How does the quadratic formula apply in non-real solutions?
5. Can this method be generalized for higher powers of \( x \)?

**Tip**: Always look for ways to simplify higher degree equations by substituting intermediate variables.

Learn how to solve the polynomial equation x^6 + x^3 - 2 = 0 using substitution, the quadratic formula, and cube roots of complex numbers. This problem involves advanced algebraic concepts suitable for Grades 11-12, including the use of complex number properties and higher-degree polynomials.

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