Solve a^2m-b^2n+a^2n-b^2m

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

Factor

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

Solution Steps

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

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

Factor out common term m+n by using distributive property.

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

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).

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

Rewrite the complete factored expression.

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

Evaluate

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