Solve 1-m^16

$$1 - m ^ { 16 }$$

Factor

$$\left(m-1\right)\left(m+1\right)\left(m^{2}+1\right)\left(m^{4}+1\right)\left(-m^{8}-1\right)$$

Solution Steps

Rewrite 1-m^{16} as 1^{2}-\left(-m^{8}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(1+m^{8}\right)\left(1-m^{8}\right)$$

Reorder the terms.

$$\left(m^{8}+1\right)\left(-m^{8}+1\right)$$

Consider -m^{8}+1. Rewrite -m^{8}+1 as 1^{2}-\left(-m^{4}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Reorder the terms.

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

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

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

Reorder the terms.

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

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

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

Reorder the terms.

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

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: m^{2}+1,m^{4}+1,m^{8}+1.

$$\left(-m+1\right)\left(m+1\right)\left(m^{2}+1\right)\left(m^{4}+1\right)\left(m^{8}+1\right)$$

Evaluate

$$1-m^{16}$$