Question

$$(a-b)^{2}-c^{2}; a^{2}-(b+c)^{2}; \frac{(b-c)^{2}-a^{2}}{b^{2}-(c+a)^{2}}+\frac{(c-a)^{2}-b^{2}}{c^{2}-(a+b)^{2}}$$

Answer

$$(a-b)^2-c^2;a^2-(b+c)^2;((b-c+a)*(b-c-a))/((b+c+a)*(b-c-a))+(c-a+b)/(c+a+b)$$

Solution


Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[\begin{aligned}&{(a-b)}^{2}-{c}^{2}\\&{a}^{2}-{(b+c)}^{2}\\&\frac{(b-c+a)(b-c-a)}{{b}^{2}-{(c+a)}^{2}}+\frac{{(c-a)}^{2}-{b}^{2}}{{c}^{2}-{(a+b)}^{2}}\end{aligned}\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[\begin{aligned}&{(a-b)}^{2}-{c}^{2}\\&{a}^{2}-{(b+c)}^{2}\\&\frac{(b-c+a)(b-c-a)}{(b+c+a)(b-(c+a))}+\frac{{(c-a)}^{2}-{b}^{2}}{{c}^{2}-{(a+b)}^{2}}\end{aligned}\]
Remove parentheses.
\[\begin{aligned}&{(a-b)}^{2}-{c}^{2}\\&{a}^{2}-{(b+c)}^{2}\\&\frac{(b-c+a)(b-c-a)}{(b+c+a)(b-c-a)}+\frac{{(c-a)}^{2}-{b}^{2}}{{c}^{2}-{(a+b)}^{2}}\end{aligned}\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[\begin{aligned}&{(a-b)}^{2}-{c}^{2}\\&{a}^{2}-{(b+c)}^{2}\\&\frac{(b-c+a)(b-c-a)}{(b+c+a)(b-c-a)}+\frac{(c-a+b)(c-a-b)}{{c}^{2}-{(a+b)}^{2}}\end{aligned}\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[\begin{aligned}&{(a-b)}^{2}-{c}^{2}\\&{a}^{2}-{(b+c)}^{2}\\&\frac{(b-c+a)(b-c-a)}{(b+c+a)(b-c-a)}+\frac{(c-a+b)(c-a-b)}{(c+a+b)(c-(a+b))}\end{aligned}\]
Remove parentheses.
\[\begin{aligned}&{(a-b)}^{2}-{c}^{2}\\&{a}^{2}-{(b+c)}^{2}\\&\frac{(b-c+a)(b-c-a)}{(b+c+a)(b-c-a)}+\frac{(c-a+b)(c-a-b)}{(c+a+b)(c-a-b)}\end{aligned}\]
Cancel \(c-a-b\).
\[\begin{aligned}&{(a-b)}^{2}-{c}^{2}\\&{a}^{2}-{(b+c)}^{2}\\&\frac{(b-c+a)(b-c-a)}{(b+c+a)(b-c-a)}+\frac{c-a+b}{c+a+b}\end{aligned}\]