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

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

Evaluate

$$2a^{2}+3b^{2}$$

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}.

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

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

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

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

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

Combine -b^{2} and 4b^{2} to get 3b^{2}.

$$2a^{2}+3b^{2}+4ab+2\left(-2\right)ab$$

Multiply 2 and -2 to get -4.

$$2a^{2}+3b^{2}+4ab-4ab$$

Combine 4ab and -4ab to get 0.

$$2a^{2}+3b^{2}$$

Expand

$$2a^{2}+3b^{2}$$

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}.

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

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

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

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

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

Combine -b^{2} and 4b^{2} to get 3b^{2}.

$$2a^{2}+3b^{2}+4ab+2\left(-2\right)ab$$

Multiply 2 and -2 to get -4.

$$2a^{2}+3b^{2}+4ab-4ab$$

Combine 4ab and -4ab to get 0.

$$2a^{2}+3b^{2}$$