Solve 1-81a^4

$$1 - 81 a ^ { 4 }$$

Factor

$$\left(3a-1\right)\left(3a+1\right)\left(-9a^{2}-1\right)$$

Solution Steps

Rewrite 1-81a^{4} as 1^{2}-\left(9a^{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).

$$\left(1-9a^{2}\right)\left(1+9a^{2}\right)$$

Reorder the terms.

$$\left(-9a^{2}+1\right)\left(9a^{2}+1\right)$$

Consider -9a^{2}+1. Rewrite -9a^{2}+1 as 1^{2}-\left(3a\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(1-3a\right)\left(1+3a\right)$$

Reorder the terms.

$$\left(-3a+1\right)\left(3a+1\right)$$

Rewrite the complete factored expression. Polynomial 9a^{2}+1 is not factored since it does not have any rational roots.

$$\left(-3a+1\right)\left(3a+1\right)\left(9a^{2}+1\right)$$

Evaluate

$$1-81a^{4}$$