The equation is:

\[
x^6 - 4x^3 - 5 = 0
\]

Let's solve it step by step.

1. We can perform a substitution to simplify the equation. Let \( y = x^3 \), then \( x^6 = (x^3)^2 = y^2 \). This transforms the original equation into:

\[
y^2 - 4y - 5 = 0
\]

2. Now, we can solve this quadratic equation. The general form of a quadratic equation is \( ay^2 + by + c = 0 \), and its solutions are given by:

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

For \( y^2 - 4y - 5 = 0 \), we have:
- \( a = 1 \)
- \( b = -4 \)
- \( c = -5 \)

Now substitute these values into the quadratic formula:

\[
y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}
\]
\[
y = \frac{4 \pm \sqrt{16 + 20}}{2}
\]
\[
y = \frac{4 \pm \sqrt{36}}{2}
\]
\[
y = \frac{4 \pm 6}{2}
\]

So, the two possible values for \( y \) are:

\[
y = \frac{4 + 6}{2} = 5 \quad \text{or} \quad y = \frac{4 - 6}{2} = -1
\]

3. Now, recall that \( y = x^3 \), so we have two equations to solve:

- \( x^3 = 5 \)
- \( x^3 = -1 \)

4. Solving these gives us the cube roots:
- \( x = \sqrt[3]{5} \)
- \( x = \sqrt[3]{-1} = -1 \)

Thus, the real solutions are \( x = \sqrt[3]{5} \) and \( x = -1 \).

Do you want more details or have any questions?

Here are five related questions for further practice:
1. How would you solve the equation \( x^6 - 4x^3 + 4 = 0 \)?
2. What are the solutions to \( x^6 - 9x^3 + 8 = 0 \)?
3. How do you solve \( x^4 - 3x^2 - 4 = 0 \)?
4. How would you approach solving \( x^3 + x - 1 = 0 \)?
5. What happens when \( y^2 + 2y + 1 = 0 \)?

**Tip**: Substitution is a helpful technique to reduce higher-degree polynomials to simpler forms, making them easier to solve.

Learn how to solve the polynomial equation x^6 - 4x^3 - 5 = 0 using the substitution method. This step-by-step guide covers key algebraic concepts, including the reduction of higher-degree polynomials to a quadratic equation, and applies the quadratic formula theorem. Suitable for students in Grades 9-12.

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