Solve a^{2}m^{2}-b^{2}m^{2}-a^2n^2+b^2n^2=

$$a ^ { 2 } m ^ { 2 } - b ^ { 2 } m ^ { 2 } - a ^ { 2 } n ^ { 2 } + b ^ { 2 } n ^ { 2 } =$$

Factor

$$\left(a-b\right)\left(m-n\right)\left(a+b\right)\left(m+n\right)$$

Solution Steps

Do the grouping a^{2}m^{2}-b^{2}m^{2}-a^{2}n^{2}+b^{2}n^{2}=\left(a^{2}m^{2}-b^{2}m^{2}\right)+\left(-a^{2}n^{2}+b^{2}n^{2}\right), and factor out m^{2} in the first and -n^{2} in the second group.

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

Factor out common term a^{2}-b^{2} by using distributive property.

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

Consider a^{2}-b^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a-b\right)\left(a+b\right)$$

Consider m^{2}-n^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

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

Rewrite the complete factored expression.

$$\left(a-b\right)\left(a+b\right)\left(m-n\right)\left(m+n\right)$$

Evaluate

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