Solve 1-a^6

$$1- { a }^{ 6 }$$

Factor

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

Solution Steps

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

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

Reorder the terms.

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

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^{3}+1. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient -1. One such root is 1. Factor the polynomial by dividing it by a-1.

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

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.

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

Evaluate

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