Solve 16a^8-b^4

$$16 a ^ { 8 } - b ^ { 4 }$$

Factor

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

Solution Steps

Rewrite 16a^{8}-b^{4} as \left(4a^{4}\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).

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

Consider 4a^{4}-b^{2}. Rewrite 4a^{4}-b^{2} as \left(2a^{2}\right)^{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).

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

Rewrite the complete factored expression.

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

Evaluate

$$16a^{8}-b^{4}$$