Solve 27b^4+81b^3-3b^2-9b

$$27 b ^ { 4 } + 81 b ^ { 3 } - 3 b ^ { 2 } - 9 b$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(9b^{4}+27b^{3}-b^{2}-3b\right)$$

Consider 9b^{4}+27b^{3}-b^{2}-3b. Factor out b.

$$b\left(9b^{3}+27b^{2}-b-3\right)$$

Consider 9b^{3}+27b^{2}-b-3. Do the grouping 9b^{3}+27b^{2}-b-3=\left(9b^{3}+27b^{2}\right)+\left(-b-3\right), and factor out 9b^{2} in the first and -1 in the second group.

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

Factor out common term b+3 by using distributive property.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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