Solve p^8-256

$$p ^ { 8 } - 256$$

Factor

$$\left(p-2\right)\left(p+2\right)\left(p^{2}+4\right)\left(p^{4}+16\right)$$

Solution Steps

Rewrite p^{8}-256 as \left(p^{4}\right)^{2}-16^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(p^{4}-16\right)\left(p^{4}+16\right)$$

Consider p^{4}-16. Rewrite p^{4}-16 as \left(p^{2}\right)^{2}-4^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(p^{2}-4\right)\left(p^{2}+4\right)$$

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

$$\left(p-2\right)\left(p+2\right)$$

Rewrite the complete factored expression. The following polynomials are not factored since they do not have any rational roots: p^{2}+4,p^{4}+16.

$$\left(p-2\right)\left(p+2\right)\left(p^{2}+4\right)\left(p^{4}+16\right)$$

Evaluate

$$p^{8}-256$$