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.