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

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

Evaluate

$$2b\left(3a-b\right)$$

Solution Steps

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}.

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

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

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

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

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

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

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

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

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

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

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

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

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

Combine 2ab and 4ab to get 6ab.

$$6ab-2b^{2}$$

Expand

$$6ab-2b^{2}$$

Solution Steps

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}.

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

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

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

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

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

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

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

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

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

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

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

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

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

Combine 2ab and 4ab to get 6ab.

$$6ab-2b^{2}$$