Solve 4a^3-4a^2-4a+4

$$4 a ^ { 3 } - 4 a ^ { 2 } - 4 a + 4$$

Factor

$$4\left(a+1\right)\left(a-1\right)^{2}$$

Solution Steps

Factor out 4.

$$4\left(a^{3}-a^{2}-a+1\right)$$

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

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

Factor out common term a-1 by using distributive property.

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

Consider a^{2}-1. Rewrite a^{2}-1 as a^{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).

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

Rewrite the complete factored expression.

$$4\left(a-1\right)^{2}\left(a+1\right)$$

Evaluate

$$4\left(a+1\right)\left(a-1\right)^{2}$$