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

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

Evaluate

$$6\left(a-3\right)$$

Solution Steps

Consider \left(a+3\right)\left(a-3\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 3.

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

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

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

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

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

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

$$-9+6a-9$$

Subtract 9 from -9 to get -18.

$$-18+6a$$

Expand

$$6a-18$$

Solution Steps

Consider \left(a+3\right)\left(a-3\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 3.

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

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

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

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

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

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

$$-9+6a-9$$

Subtract 9 from -9 to get -18.

$$-18+6a$$