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

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

Evaluate

$$0$$

Solution Steps

Use the distributive property to multiply a^{2}+1 by b^{2}+1.

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

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(ab-1\right)^{2}.

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

To find the opposite of a^{2}b^{2}-2ab+1, find the opposite of each term.

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

Combine a^{2}b^{2} and -a^{2}b^{2} to get 0.

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

Subtract 1 from 1 to get 0.

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

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+b\right)^{2}.

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

To find the opposite of a^{2}+2ab+b^{2}, find the opposite of each term.

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

Combine a^{2} and -a^{2} to get 0.

$$b^{2}+2ab-2ab-b^{2}$$

Combine 2ab and -2ab to get 0.

$$b^{2}-b^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$0$$

Factor

$$0$$