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

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

Evaluate

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

Solution Steps

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

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

Use the distributive property to multiply 3 by a^{2}+2ab+b^{2}.

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

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

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

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

$$3a^{2}+6ab+3b^{2}-a^{2}+2ab-b^{2}$$

Combine 3a^{2} and -a^{2} to get 2a^{2}.

$$2a^{2}+6ab+3b^{2}+2ab-b^{2}$$

Combine 6ab and 2ab to get 8ab.

$$2a^{2}+8ab+3b^{2}-b^{2}$$

Combine 3b^{2} and -b^{2} to get 2b^{2}.

$$2a^{2}+8ab+2b^{2}$$

Expand

$$2a^{2}+8ab+2b^{2}$$

Solution Steps

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

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

Use the distributive property to multiply 3 by a^{2}+2ab+b^{2}.

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

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

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

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

$$3a^{2}+6ab+3b^{2}-a^{2}+2ab-b^{2}$$

Combine 3a^{2} and -a^{2} to get 2a^{2}.

$$2a^{2}+6ab+3b^{2}+2ab-b^{2}$$

Combine 6ab and 2ab to get 8ab.

$$2a^{2}+8ab+3b^{2}-b^{2}$$

Combine 3b^{2} and -b^{2} to get 2b^{2}.

$$2a^{2}+8ab+2b^{2}$$