Solve (16a^4-b^4)

$$( 16 a ^ { 4 } - b ^ { 4 } )$$

Factor

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

Solution Steps

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

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

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

$$\left(2a-b\right)\left(2a+b\right)$$

Rewrite the complete factored expression.

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

Evaluate

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