Solve (m+n)^2+(m+n)(m-n)

$$( m + n ) ^ { 2 } + ( m + n ) ( m - n )$$

Evaluate

$$2m\left(m+n\right)$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(m+n\right)^{2}.

$$m^{2}+2mn+n^{2}+\left(m+n\right)\left(m-n\right)$$

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

$$m^{2}+2mn+n^{2}+m^{2}-n^{2}$$

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

$$2m^{2}+2mn+n^{2}-n^{2}$$

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

$$2m^{2}+2mn$$

Expand

$$2m^{2}+2mn$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(m+n\right)^{2}.

$$m^{2}+2mn+n^{2}+\left(m+n\right)\left(m-n\right)$$

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

$$m^{2}+2mn+n^{2}+m^{2}-n^{2}$$

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

$$2m^{2}+2mn+n^{2}-n^{2}$$

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

$$2m^{2}+2mn$$