The problem asks us to factor the expression $25y^6 - 80y^3 + 64$ completely.

1. Problem Description

The problem asks us to factor the expression $25y^6 - 80y^3 + 64$ completely.

2. Solution Steps

Let $x = y^3$. Then the expression becomes $25x^2 - 80x + 64$.

This is a quadratic expression in $x$. We can try to factor it.

We are looking for two numbers that multiply to $25 \times 64 = 1600$ and add up to $-80$.

The numbers are $-40$ and $-40$.

So we can rewrite the middle term as $-40x - 40x$:

$25x^2 - 40x - 40x + 64$.

Now we can factor by grouping:

$5x(5x - 8) - 8(5x - 8) = (5x - 8)(5x - 8) = (5x - 8)^2$.

Now substitute $x = y^3$ back into the expression:

$(5y^3 - 8)^2$.

We can write $8 = 2^3$, so we have $(5y^3 - 2^3)^2$.

However, we can not factor further in real numbers.

We can also try to recognize the quadratic expression as a perfect square trinomial:

$25x^2 - 80x + 64 = (5x)^2 - 2(5x)(8) + (8)^2 = (5x - 8)^2$.

Substitute $x = y^3$:

$(5y^3 - 8)^2$.

3. Final Answer

$(5y^3-8)^2$