Solve 16a^4-81b^4

$$16 a ^ { 4 } - 81 b ^ { 4 }$$

Factor

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

Solution Steps

Rewrite 16a^{4}-81b^{4} as \left(4a^{2}\right)^{2}-\left(9b^{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).

$$\left(4a^{2}-9b^{2}\right)\left(4a^{2}+9b^{2}\right)$$

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

$$\left(2a-3b\right)\left(2a+3b\right)$$

Rewrite the complete factored expression.

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

Evaluate

$$16a^{4}-81b^{4}$$