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.