Solve a^{20}-a^{16}+a^12-a^6+a^4-a^2

$$a ^ { 20 } - a ^ { 16 } + a ^ { 12 } - a ^ { 6 } + a ^ { 4 } - a ^ { 2 }$$

Factor

$$\left(a-1\right)\left(a+1\right)a^{2}\left(a^{2}-a+1\right)\left(a^{2}+a+1\right)\left(a^{12}-a^{8}+a^{6}+a^{4}-a^{2}+1\right)$$

Solution Steps

Factor out a^{2}.

$$a^{2}\left(a^{18}-a^{14}+a^{10}-a^{4}+a^{2}-1\right)$$

Consider a^{18}-a^{14}+a^{10}-a^{4}+a^{2}-1. Find one factor of the form a^{k}+m, where a^{k} divides the monomial with the highest power a^{18} and m divides the constant factor -1. One such factor is a^{6}-1. Factor the polynomial by dividing it by this factor.

$$\left(a^{6}-1\right)\left(a^{12}-a^{8}+a^{6}+a^{4}-a^{2}+1\right)$$

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).

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

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^{12}-a^{8}+a^{6}+a^{4}-a^{2}+1.

$$a^{2}\left(a-1\right)\left(a^{2}+a+1\right)\left(a+1\right)\left(a^{2}-a+1\right)\left(a^{12}-a^{8}+a^{6}+a^{4}-a^{2}+1\right)$$

Evaluate

$$a^{2}\left(a^{18}-a^{14}+a^{10}-a^{4}+a^{2}-1\right)$$