Solve A^6-1

$$A ^ { 6 } - 1$$

Factor

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

Solution Steps

Rewrite A^{6}-1 as \left(A^{3}\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).

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

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

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

Evaluate

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