To solve the equation:

\[
x^6 + x^3 - 4 = 0
\]

we can proceed by making a substitution that simplifies it. Let:

\[
y = x^3
\]

Then \(x^6 = (x^3)^2 = y^2\). Substitute into the equation:

\[
y^2 + y - 4 = 0
\]

Now, we have a quadratic equation in terms of \(y\):

\[
y^2 + y - 4 = 0
\]

To solve this, we can use the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 1\), \(b = 1\), and \(c = -4\):

\[
y = \frac{-1 \pm \sqrt{1 + 16}}{2} = \frac{-1 \pm \sqrt{17}}{2}
\]

This gives two solutions for \(y\):

\[
y = \frac{-1 + \sqrt{17}}{2} \quad \text{and} \quad y = \frac{-1 - \sqrt{17}}{2}
\]

Now we need to revert back to \(x\) by substituting \(y = x^3\):

1. For \(y = \frac{-1 + \sqrt{17}}{2}\):
   \[
   x^3 = \frac{-1 + \sqrt{17}}{2}
   \]

2. For \(y = \frac{-1 - \sqrt{17}}{2}\):
   \[
   x^3 = \frac{-1 - \sqrt{17}}{2}
   \]

Now, we would take the cube root of each expression to find the values of \(x\):

\[
x = \sqrt[3]{\frac{-1 + \sqrt{17}}{2}} \quad \text{and} \quad x = \sqrt[3]{\frac{-1 - \sqrt{17}}{2}}
\]

Thus, the solutions are:

\[
x = \sqrt[3]{\frac{-1 + \sqrt{17}}{2}} \quad \text{and} \quad x = \sqrt[3]{\frac{-1 - \sqrt{17}}{2}}
\]

Let me know if you would like a further breakdown of these steps or have any questions.

### Related Questions

1. What happens if we solve similar equations with higher powers, like \(x^8 + x^4 - 4 = 0\)?
2. How can we graph the function \(x^6 + x^3 - 4\) to visualize the roots?
3. Are there general methods for solving equations of the form \(x^{2n} + x^n - c = 0\)?
4. What are the complex solutions of this equation if complex numbers are allowed?
5. How does changing the constant from \(4\) to another number affect the solution?

#### Tip:
When solving higher-degree polynomials, substituting \(y = x^n\) can often reduce the problem to a simpler quadratic form.

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