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

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

Evaluate

$$6a$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(3a+1\right)^{2}.

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

Consider \left(3a-1\right)\left(3a+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}. Square 1.

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

Expand \left(3a\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

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

To find the opposite of 9a^{2}-1, find the opposite of each term.

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

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

$$6a+1+1-2$$

Add 1 and 1 to get 2.

$$6a+2-2$$

Subtract 2 from 2 to get 0.

$$6a$$

Expand

$$6a$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(3a+1\right)^{2}.

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

Consider \left(3a-1\right)\left(3a+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}. Square 1.

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

Expand \left(3a\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

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

To find the opposite of 9a^{2}-1, find the opposite of each term.

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

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

$$6a+1+1-2$$

Add 1 and 1 to get 2.

$$6a+2-2$$

Subtract 2 from 2 to get 0.

$$6a$$