The problem asks us to factor the expression $(x^2-1)(x^3-1)(2x^2-x-1)$.

1. Problem Description

The problem asks us to factor the expression $(x^2-1)(x^3-1)(2x^2-x-1)$.

2. Solution Steps

First, we factor each term in the expression.

$x^2 - 1$ can be factored using the difference of squares formula:

$a^2 - b^2 = (a-b)(a+b)$

So, $x^2 - 1 = (x-1)(x+1)$.

Next, we factor $x^3 - 1$. This is a difference of cubes, which factors as:

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

Thus, $x^3 - 1 = (x-1)(x^2 + x + 1)$.

Finally, we factor $2x^2 - x - 1$. We are looking for two numbers that multiply to $(2)(-1) = -2$ and add up to $-1$. Those two numbers are $-2$ and $1$. We rewrite the middle term using these numbers:

$2x^2 - x - 1 = 2x^2 - 2x + x - 1$

Now we factor by grouping:

$2x^2 - 2x + x - 1 = 2x(x-1) + 1(x-1) = (2x+1)(x-1)$.

Now, we can rewrite the original expression using the factored forms:

$(x^2-1)(x^3-1)(2x^2-x-1) = (x-1)(x+1)(x-1)(x^2+x+1)(x-1)(2x+1)$

$= (x-1)^3 (x+1) (x^2+x+1) (2x+1)$.

3. Final Answer

$(x-1)^3(x+1)(x^2+x+1)(2x+1)$