Solve {l}{(a^{m}+a^{n})^{2}-(a^{m}+a^{n})(a^{m}-a^{n})-2a^n(a^m+a^n)}{0^2}

$$\left. \begin{array} { l } { ( a ^ { m } + a ^ { n } ) ^ { 2 } - ( a ^ { m } + a ^ { n } ) ( a ^ { m } - a ^ { n } ) - 2 a ^ { n } ( a ^ { m } + a ^ { n } ) } \\ { 0 ^ { 2 } } \end{array} \right.$$

Sort

$$0,\ 0$$

Solution Steps

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

$$sort(\left(a^{m}\right)^{2}+2a^{m}a^{n}+\left(a^{n}\right)^{2}-\left(a^{m}+a^{n}\right)\left(a^{m}-a^{n}\right)-2a^{n}\left(a^{m}+a^{n}\right),0^{2})$$

Consider \left(a^{m}+a^{n}\right)\left(a^{m}-a^{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}.

$$sort(\left(a^{m}\right)^{2}+2a^{m}a^{n}+\left(a^{n}\right)^{2}-\left(\left(a^{m}\right)^{2}-\left(a^{n}\right)^{2}\right)-2a^{n}\left(a^{m}+a^{n}\right),0^{2})$$

Calculate 0 to the power of 2 and get 0.

$$sort(\left(a^{m}\right)^{2}+2a^{m}a^{n}+\left(a^{n}\right)^{2}-\left(\left(a^{m}\right)^{2}-\left(a^{n}\right)^{2}\right)-2a^{n}\left(a^{m}+a^{n}\right),0)$$

To find the opposite of \left(a^{m}\right)^{2}-\left(a^{n}\right)^{2}, find the opposite of each term.

$$sort(\left(a^{m}\right)^{2}+2a^{m}a^{n}+\left(a^{n}\right)^{2}-\left(a^{m}\right)^{2}+\left(a^{n}\right)^{2}-2a^{n}\left(a^{m}+a^{n}\right),0)$$

Combine \left(a^{m}\right)^{2} and -\left(a^{m}\right)^{2} to get 0.

$$sort(2a^{m}a^{n}+\left(a^{n}\right)^{2}+\left(a^{n}\right)^{2}-2a^{n}\left(a^{m}+a^{n}\right),0)$$

Combine \left(a^{n}\right)^{2} and \left(a^{n}\right)^{2} to get 2\left(a^{n}\right)^{2}.

$$sort(2a^{m}a^{n}+2\left(a^{n}\right)^{2}-2a^{n}\left(a^{m}+a^{n}\right),0)$$

Use the distributive property to multiply -2a^{n} by a^{m}+a^{n}.

$$sort(2a^{m}a^{n}+2\left(a^{n}\right)^{2}-2a^{n}a^{m}-2\left(a^{n}\right)^{2},0)$$

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

$$sort(2\left(a^{n}\right)^{2}-2\left(a^{n}\right)^{2},0)$$

Combine 2\left(a^{n}\right)^{2} and -2\left(a^{n}\right)^{2} to get 0.

$$sort(0,0)$$

The list values are already in order.

$$0,0$$

Evaluate

$$0,\ 0$$