Solve a^2b^2+2ab+1-a^2-2ab-b^2

$$a ^ { 2 } b ^ { 2 } + 2 a b + 1 - a ^ { 2 } - 2 a b - b ^ { 2 }$$

Factor

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

Solution Steps

Multiply and combine like terms.

$$a^{2}b^{2}-a^{2}-b^{2}+1$$

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

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

Factor out common term b^{2}-1 by using distributive property.

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

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

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

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.

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

Evaluate

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