Solve (m+1)(m-1)=m^2-1

$$( m + 1 ) ( m - 1 ) = m ^ { 2 } - 1$$

Solve for m (complex solution)

$$m\in \mathrm{C}$$

Solution Steps

Consider \left(m+1\right)\left(m-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$m^{2}-1=m^{2}-1$$

Subtract m^{2} from both sides.

$$m^{2}-1-m^{2}=-1$$

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

$$-1=-1$$

Compare -1 and -1.

$$\text{true}$$

This is true for any m.

$$m\in \mathrm{C}$$

Solve for m

$$m\in \mathrm{R}$$

Solution Steps

Consider \left(m+1\right)\left(m-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$m^{2}-1=m^{2}-1$$

Subtract m^{2} from both sides.

$$m^{2}-1-m^{2}=-1$$

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

$$-1=-1$$

Compare -1 and -1.

$$\text{true}$$

This is true for any m.

$$m\in \mathrm{R}$$