To verify the factorization \( n^4 + n^2 + 1 = (n^2 - n + 1)(n^2 + n + 1) \), we'll expand the right-hand side and check if it equals the left-hand side.

1. Start with the right-hand side:
\[
(n^2 - n + 1)(n^2 + n + 1)
\]

2. Apply the distributive property (also known as the FOIL method):
- Multiply \( n^2 \) by each term in \( (n^2 + n + 1) \):
\[
n^2 \cdot n^2 = n^4
\]
\[
n^2 \cdot n = n^3
\]
\[
n^2 \cdot 1 = n^2
\]

- Multiply \( -n \) by each term in \( (n^2 + n + 1) \):
\[
-n \cdot n^2 = -n^3
\]
\[
-n \cdot n = -n^2
\]
\[
-n \cdot 1 = -n
\]

- Multiply \( 1 \) by each term in \( (n^2 + n + 1) \):
\[
1 \cdot n^2 = n^2
\]
\[
1 \cdot n = n
\]
\[
1 \cdot 1 = 1
\]

3. Now collect all of these results:
\[
n^4 + n^3 + n^2 - n^3 - n^2 - n + n^2 + n + 1
\]

4. Combine like terms:
\[
n^4 + (n^3 - n^3) + (n^2 - n^2 + n^2) + (-n + n) + 1 = n^4 + 0 + n^2 + 0 + 1
\]
Which simplifies to:
\[
n^4 + n^2 + 1
\]

Since the right-hand side expands to \( n^4 + n^2 + 1 \), we can conclude:
\[
n^4 + n^2 + 1 = (n^2 - n + 1)(n^2 + n + 1)
\]
is indeed correct. Thus, the factorization holds true.