Answer:
[tex]\displaystyle \frac{(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} }=1[/tex]
Step-by-step explanation:
Algebra Simplifying
We are given the expression:
[tex]\displaystyle T=\frac{(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} }[/tex]
We need to repeatedly use the following identity:
[tex](a^2-b^2)=(a-b)(a+b)[/tex]
For example, the first numerator has a difference of squares thus we factor as:
[tex](a - b)^{2} - {c}^{2} =(a-b-c)(a-b+c)[/tex]
The first denominator can be factored also:
[tex]{a}^{2} - {(b+ c)}^{2} = (a-b-c)(a+b+c)[/tex]
Applying the same procedure to all the expressions:
[tex]\displaystyle T=\frac{(a - b- c)(a-b+c) }{a-b-c)(a+b+c) } + \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)}[/tex]
Simplifying all the fractions:
[tex]\displaystyle T=\frac{a - b- c}{a+b+c} + \frac{b-c+a}{b+c+a } + \frac{c-a+b }{c+a+b}[/tex]
Since all the denominators are equal:
[tex]\displaystyle T=\frac{a - b+ c+b-c+a+c-a+b}{a+b+c}[/tex]
Simplifying:
[tex]\displaystyle T=\frac{a +c+b}{a+b+c}[/tex]
Simplifying again:
T = 1
Thus:
[tex]\boxed{\displaystyle \frac{(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} }=1}[/tex]
