Solve 81n^4-1

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

Factor

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

Solution Steps

Rewrite 81n^{4}-1 as \left(9n^{2}\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

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

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

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

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

Evaluate

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