Solve a^{20}-a^{16}+a^2-a^8+a^4-a^2

$$a ^ { 20 } - a ^ { 16 } + a ^ { 2 } - a ^ { 8 } + a ^ { 4 } - a ^ { 2 }$$

Factor

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

Solution Steps

Factor out a^{2}.

$$a^{2}\left(a^{2}+a^{18}-a^{6}-a^{14}\right)$$

Consider a^{18}-a^{14}+1-a^{6}+a^{2}-1. Multiply and combine like terms.

$$a^{18}-a^{14}-a^{6}+a^{2}$$

Consider a^{18}-a^{14}-a^{6}+a^{2}. Factor out a^{2}.

$$a^{2}\left(a^{16}-a^{12}-a^{4}+1\right)$$

Consider a^{16}-a^{12}-a^{4}+1. Do the grouping a^{16}-a^{12}-a^{4}+1=\left(a^{16}-a^{12}\right)+\left(-a^{4}+1\right), and factor out a^{12} in the first and -1 in the second group.

$$a^{12}\left(a^{4}-1\right)-\left(a^{4}-1\right)$$

Factor out common term a^{4}-1 by using distributive property.

$$\left(a^{4}-1\right)\left(a^{12}-1\right)$$

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

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

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

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

Consider a^{12}-1. Rewrite a^{12}-1 as \left(a^{6}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a^{6}-1\right)\left(a^{6}+1\right)$$

Consider a^{6}-1. Rewrite a^{6}-1 as \left(a^{3}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a^{3}-1\right)\left(a^{3}+1\right)$$

Consider a^{3}-1. Rewrite a^{3}-1 as a^{3}-1^{3}. The difference of cubes can be factored using the rule: p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right).

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

Consider a^{3}+1. Rewrite a^{3}+1 as a^{3}+1^{3}. The sum of cubes can be factored using the rule: p^{3}+q^{3}=\left(p+q\right)\left(p^{2}-pq+q^{2}\right).

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

Consider a^{6}+1. Rewrite a^{6}+1 as \left(a^{2}\right)^{3}+1^{3}. The sum of cubes can be factored using the rule: p^{3}+q^{3}=\left(p+q\right)\left(p^{2}-pq+q^{2}\right).

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

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: a^{2}-a+1,a^{2}+a+1,a^{4}-a^{2}+1,a^{2},a^{2}+1.

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

Evaluate

$$a^{20}-a^{16}-a^{8}+a^{4}$$

Solution Steps

Combine a^{2} and -a^{2} to get 0.

$$a^{20}-a^{16}-a^{8}+a^{4}$$