Solve n^6-n^4-n^2+1

$$n ^ { 6 } - n ^ { 4 } - n ^ { 2 } + 1$$

Factor

$$\left(n^{2}+1\right)\left(n^{2}-1\right)^{2}$$

Solution Steps

Do the grouping n^{6}-n^{4}-n^{2}+1=\left(n^{6}-n^{4}\right)+\left(-n^{2}+1\right), and factor out n^{4} in the first and -1 in the second group.

$$n^{4}\left(n^{2}-1\right)-\left(n^{2}-1\right)$$

Factor out common term n^{2}-1 by using distributive property.

$$\left(n^{2}-1\right)\left(n^{4}-1\right)$$

Consider n^{2}-1. Rewrite n^{2}-1 as n^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(n-1\right)\left(n+1\right)$$

Consider n^{4}-1. Rewrite n^{4}-1 as \left(n^{2}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(n^{2}-1\right)\left(n^{2}+1\right)$$

Consider n^{2}-1. Rewrite n^{2}-1 as n^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(n-1\right)\left(n+1\right)$$

Rewrite the complete factored expression. Polynomial n^{2}+1 is not factored since it does not have any rational roots.

$$\left(n^{2}+1\right)\left(n-1\right)^{2}\left(n+1\right)^{2}$$

Evaluate

$$\left(n^{2}+1\right)\left(n^{2}-1\right)^{2}$$