Question

Plz help with full process

Answer 1

Answer:

$\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1$

Step-by-step explanation:

Algebra Simplifying

We are given the expression:

$\displaystyle T=\frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }$

We need to repeatedly use the following identity:

$(a^2-b^2)=(a-b)(a+b)$

For example, the first numerator has a difference of squares thus we factor as:

$(a - b)^{2} - {c}^{2} =(a-b-c)(a-b+c)$

The first denominator can be factored also:

${a}^{2} - {(b+ c)}^{2} = (a-b-c)(a+b+c)$

Applying the same procedure to all the expressions:

$\displaystyle T=\frac{(a - b- c)(a-b+c) }{a-b-c)(a+b+c) } + \frac{(b - c - a)(b-c+a) }{(b-c-a)(b+c+a) } + \frac{(c - a-b)(c-a+b) }{(c-a-b)(c+a+b)}$

Simplifying all the fractions:

$\displaystyle T=\frac{a - b- c}{a+b+c} + \frac{b-c+a}{b+c+a } + \frac{c-a+b }{c+a+b}$

Since all the denominators are equal:

$\displaystyle T=\frac{a - b+ c+b-c+a+c-a+b}{a+b+c}$

Simplifying:

$\displaystyle T=\frac{a +c+b}{a+b+c}$

Simplifying again:

T = 1

Thus:

$\boxed{\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1}$

Answer 2

Answer:

Siblings r annoying

Step-by-step explanation: