Solve n^8-81

$$n ^ { 8 } - 81$$

Factor

$$\left(n^{2}-3\right)\left(n^{2}+3\right)\left(n^{4}+9\right)$$

Solution Steps

Rewrite n^{8}-81 as \left(n^{4}\right)^{2}-9^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(n^{4}-9\right)\left(n^{4}+9\right)$$

Consider n^{4}-9. Rewrite n^{4}-9 as \left(n^{2}\right)^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(n^{2}-3\right)\left(n^{2}+3\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: n^{2}-3,n^{2}+3,n^{4}+9.

$$\left(n^{2}-3\right)\left(n^{2}+3\right)\left(n^{4}+9\right)$$

Evaluate

$$n^{8}-81$$