The given equation is:

$x^6 - 5x^3 - 6 = 0$

To solve this equation, we can use a substitution method by letting:

$y = x^3$

This substitution simplifies the equation to:

$y^2 - 5y - 6 = 0$

Now, this is a standard quadratic equation, which we can solve using the quadratic formula:

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

where $a = 1$, $b = -5$, and $c = -6$. Plugging in these values:

$y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-6)}}{2(1)}$

$y = \frac{5 \pm \sqrt{25 + 24}}{2}$

$y = \frac{5 \pm \sqrt{49}}{2}$

$y = \frac{5 \pm 7}{2}$

This gives us two solutions for $y$:

$y = \frac{5 + 7}{2} = 6 \quad \text{and} \quad y = \frac{5 - 7}{2} = -1$

Now, recall that $y = x^3$. So we have:

1. $x^3 = 6$
2. $x^3 = -1$

Finding $x$:

1. For $x^3 = 6$:

   $x = \sqrt[3]{6}$

2. For $x^3 = -1$:

   $x = \sqrt[3]{-1} = -1$

Final Real Solutions:

$x = \sqrt[3]{6}, \, -1$

So the real solutions are:

$\boxed{\sqrt[3]{6}, -1}$

Would you like further details or have any questions?

Expanding Questions:

1. How do you solve similar polynomial equations with higher powers?
2. What is the significance of using substitution in solving equations?
3. How can complex solutions be identified in similar equations?
4. How does the discriminant affect the nature of the solutions in a quadratic equation?
5. Can this method be extended to equations with different forms?

Tip: Always try substitution when facing equations with a mix of different powers, as it can greatly simplify the problem.