Hello! That's quite an interesting expression you got there of \((a-7)^2-2(a-7)(a+7)+a+7\). Let's simplify it the we can.
\((a-7)^2-2(a-7)(a+7)+a+7\) | First, let's deal with \((a-7)^2\) by knowing the expansion of a binomial squared. In general, it is \((x-y)^2=x^2-2xy+y^2\). |
\((a-7)^2=a^2-14a+49\) | |
\(a^2-14a+49-2(a-7)(a+7)+a+7\) | Now, let's expand \((a-7)(a+7)\) by using the following rule again of \((x+y)(x-y)=x^2-y^2\). |
\(-2(a-7)(a+7)=-2(a^2-49)\) | Distribute the -2 to both terms in the parentheses. |
\(-2(a^2-49)=-2a^2+98\) | |
\(a^2-14a+49-2a^2+98+a+7\) | Let's rearrange the equation such that all the terms with the same degree are adjacent. |
\(-2a^2+a^2-14a+a+98+49+7\) | Now, simplify. |
\(-a^2-13a+154\) | |
Therefore, \((a-7)^2-2(a-7)(a+7)+a+7=-a^2-13a+154\).