Find the LCM x^3-x^2+x-1 , x^2-1

$x^{3} - x^{2} + x - 1$ , $x^{2} - 1$

Step 1

Factor $x^{3} - x^{2} + x - 1$ .

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x - 1) + 1 (x - 1)$

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$(x - 1) (x^{2} + 1)$

Step 2

Factor $x^{2} - 1$ .

Rewrite $1$ as $1^{2}$ .

$x^{2} - 1^{2}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$(x + 1) (x - 1)$

Step 3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 5

The LCM of $1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 6

The factor for $x - 1$ is $x - 1$ itself.

$(x - 1) = x - 1$

$(x - 1)$ occurs $1$ time.

Step 7

The factor for $x^{2} + 1$ is $x^{2} + 1$ itself.

$(x^{2} + 1) = x^{2} + 1$

$(x^{2} + 1)$ occurs $1$ time.

Step 8

The factor for $x + 1$ is $x + 1$ itself.

$(x + 1) = x + 1$

$(x + 1)$ occurs $1$ time.

Step 9

The factor for $x - 1$ is $x - 1$ itself.

$(x - 1) = x - 1$

$(x - 1)$ occurs $1$ time.

Step 10

The LCM of $x - 1, x^{2} + 1, x + 1, x - 1$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x - 1) (x^{2} + 1) (x + 1)$