Solve b-81b^5

$$b - 81 b ^ { 5 }$$

Factor

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

Solution Steps

Factor out b.

$$b\left(1-81b^{4}\right)$$

Consider 1-81b^{4}. Rewrite 1-81b^{4} as 1^{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).

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

Reorder the terms.

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

Consider -9b^{2}+1. Rewrite -9b^{2}+1 as 1^{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).

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

Reorder the terms.

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

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

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

Evaluate

$$b-81b^{5}$$