Solve a^8-b^8

$${ a }^{ 8 } - { b }^{ 8 }$$

Factor

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

Solution Steps

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

$$\left(a^{4}-b^{4}\right)\left(a^{4}+b^{4}\right)$$

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

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

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

$$\left(a-b\right)\left(a+b\right)$$

Rewrite the complete factored expression.

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

Evaluate

$$a^{8}-b^{8}$$