Solve (a+3b-1)(a+3b+1)-(a+3b)^2-2

$$( a + 3 b - 1 ) ( a + 3 b + 1 ) - ( a + 3 b ) ^ { 2 } - 2$$

Evaluate

$$-3$$

Solution Steps

Consider \left(a+3b-1\right)\left(a+3b+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}, where a=a+3b and b=1. Square 1.

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

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+3b\right)^{2}.

$$a^{2}+6ab+9b^{2}-1-\left(a+3b\right)^{2}-2$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+3b\right)^{2}.

$$a^{2}+6ab+9b^{2}-1-\left(a^{2}+6ab+9b^{2}\right)-2$$

To find the opposite of a^{2}+6ab+9b^{2}, find the opposite of each term.

$$a^{2}+6ab+9b^{2}-1-a^{2}-6ab-9b^{2}-2$$

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

$$6ab+9b^{2}-1-6ab-9b^{2}-2$$

Combine 6ab and -6ab to get 0.

$$9b^{2}-1-9b^{2}-2$$

Combine 9b^{2} and -9b^{2} to get 0.

$$-1-2$$

Subtract 2 from -1 to get -3.

$$-3$$

Factor

$$-3$$