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

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

Evaluate

$$a^{2}+4ab-b^{2}$$

Solution Steps

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

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

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

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

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

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

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

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

Combine 2ab and 2ab to get 4ab.

$$4ab+b^{2}-b^{2}+\left(a+b\right)\left(a-b\right)$$

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

$$4ab+\left(a+b\right)\left(a-b\right)$$

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

$$4ab+a^{2}-b^{2}$$

Expand

$$a^{2}+4ab-b^{2}$$

Solution Steps

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

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

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

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

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

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

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

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

Combine 2ab and 2ab to get 4ab.

$$4ab+b^{2}-b^{2}+\left(a+b\right)\left(a-b\right)$$

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

$$4ab+\left(a+b\right)\left(a-b\right)$$

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

$$4ab+a^{2}-b^{2}$$