Solve for \(x\) in \(x^4 - 13x^2 + 36 = 0\).

First start by converting this trinomial into a form that is more common. Here, it would be a lot easier when factoring \(x^2 - 13x + 36 = 0.\) There is a standard strategy to achieve this through substitution.

First, let \(u = x^2\). Now substitute \(u\) for every \(x^2\), the equation is transformed into \(u^2-13u+36=0\).

\(u^2 - 13u + 36 = 0\) factors into \((u - 9)(u - 4) = 0\).

Once the equation is factored, replace the substitutions with the original variables, which means that, since \(u = x^2\), then \((u - 9)(u - 4) = 0\) becomes \((x^2 - 9)(x^2 - 4) = 0\).

To complete the factorization and find the solutions for \(x\), then \((x^2 - 9)(x^2 - 4) = 0\) must be factored once more. This is done using the difference of squares equation: \(a^2 - b^2 = (a + b)(a - b)\).

Factoring \((x^2 - 9)(x^2 - 4) = 0\) thus leaves  \((x - 3)(x + 3)(x - 2)(x + 2) = 0\).

Solving each of these terms yields the solutions \(x = \pm 3, \pm 2\).

Factor the binomial \(x^6 - 7x^3 - 8 = 0\).

Here, it would be a lot easier if the expression for factoring was \(x^2 - 7x - 8 = 0\).

First, let \(u = x^3\), which leaves the factor of \(u^2 - 7u - 8 = 0\).

\(u^2 - 7u - 8 = 0\) easily factors out to \((u - 8)(u + 1) = 0\).

Now that  the substituted values are factored out, replace the \(u\) with the original \(x^3\). This turns \((u - 8)(u + 1) = 0\) into \((x^3 - 8)(x^3 + 1) = 0\).

The factored \((x^3 - 8)\) and \((x^3 + 1)\) terms can be recognized as the difference of cubes.

These are factored using \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\) and \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\).

And so, \((x^3 - 8)\) factors out to \((x - 2)(x^2 + 2x + 4)\) and \((x^3 + 1)\) factors out to \((x + 1)(x^2 - x + 1)\).

Combining all of these terms yields:

\[(x - 2)(x^2 + 2x + 4)(x + 1)(x^2 - x + 1) = 0\]

The two real solutions are \(x = 2\) and \(x = -1\). Checking for any others by using the discriminant reveals that all other solutions are complex or imaginary solutions.