Simplify: $\dfrac{(a-b)^2-c^2}{a^2-(b+c)^2} + \dfrac{(b-c)^2-a^2}{b^2-(c+a)^2} +\dfrac{(c-a)^2+b^2}{c^2-(a+b)^2}$

This is a class 10 Question From Simplification of Rational Expressions chapter of Unit Algebra (Mathematics). All the steps for the solutions are mentioned in the description below. If that's hard for you to navigate, you can always visit the facebook link given at the end of every posts. 

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Solution:

Given,

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

= $\dfrac{(a-b+c)(a-b-c)}{(a+b+c)(a-b-c)} + \dfrac{(b-c+a)(b-c-a)}{(b+c+a)(b-c-a)} +\dfrac{(c-a+b)(c-a-b)}{(c+a+b)(c-a-b)}$

= $\dfrac{a-b+c}{a+b+c} + \dfrac{b-c+a}{b+c+a} +\dfrac{c-a+b}{c+a+b}$

= $\dfrac{a-b+c+b-c+a+c-a+b}{a+b+c}$

= $\dfrac{a+b+c}{a+b+c}$

= 1

= Answer


Explanation to the above answer.


Step 1: Copy the same question given.

Step 2: Every terms either in the numerator or the denominator are in the form of a²-b². So, we expand them as the factor (a+b)(a-b).

Step 3: Cancel the common terms in the numerator as well as denominator of the same term. 

Step 4: As we have same denominators in all of the terms, we add all the numerators.

Step 5: Similar terms with opposite signs get cancelled out; eg. +a-a = 0, and we get (a+b+c) in the numerator as well as denominator.

Step 6: When we divide the one expression by the same expression, we get the result '1'.


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Question: Simplify: {(a-b)²-c²}/{a²-(b+c)²} + {(b-c)²-a²}{b²-(c+a)²} + {(c-a)²+b²}{c²-(a+b)²} | SciPiPupil

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