The problem asks to factor the expression $125m^6 + 8$ completely.

1. Problem Description

The problem asks to factor the expression $125m^6 + 8$ completely.

2. Solution Steps

We can recognize this as a sum of cubes. We have $125m^6 = (5m^2)^3$ and $8 = 2^3$. Thus, we have the expression $(5m^2)^3 + 2^3$.

The sum of cubes factorization formula is:

$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$.

In our case, $a = 5m^2$ and $b = 2$. Plugging these values into the formula, we have:

$(5m^2)^3 + 2^3 = (5m^2 + 2)((5m^2)^2 - (5m^2)(2) + 2^2) = (5m^2 + 2)(25m^4 - 10m^2 + 4)$.

Now we need to determine if $25m^4 - 10m^2 + 4$ can be factored further.

Let $x = m^2$. Then the quadratic becomes $25x^2 - 10x + 4$. The discriminant is $(-10)^2 - 4(25)(4) = 100 - 400 = -300$. Since the discriminant is negative, the quadratic $25x^2 - 10x + 4$ has no real roots, which means $25m^4 - 10m^2 + 4$ cannot be factored further using real numbers.

3. Final Answer

$(5m^2 + 2)(25m^4 - 10m^2 + 4)$