Question

Sorol koro b+c/bc(b+c-a)+ c+a/ca(c+a-b)+a+b/ab(a+b-c)

Answer 1

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Step-by-step explanation:

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Answer 2

Explanation:

• Given equation=$\frac{b+c(b+c-a)}{bc}+\frac{c+a(c+a-b)}{ca}+\frac{a+b(a+b-c)}{ab}$
• We will solve the given equation by simply multiplying .

Steps:

Step 1 of 2:

$\frac{b+c(b+c-a)}{bc}=\frac{b^2+bc-ab+bc+c^2-ac}{bc}=\frac{b^2+c^2+2bc-ab-ac}{bc}$

$\frac{c+a(c+a-b)}{ca}=\frac{c^2+ac-bc+ac+a^2-ab}{ca}=\frac{c^2+a^2+2ac-bc-ab}{ca}$

$\frac{a+b(a+b-c)}{ab}=\frac{a^2+ab-ac+ab+b^2-bc}{ab}=\frac{a^2+b^2+2ab-ac-bc}{ab}$

Step 2 of 2:

$\frac{b+c(b+c-a)}{bc}+\frac{c+a(c+a-b)}{ca}+\frac{a+b(a+b-c)}{ab}=\frac{b^2+c^2+2bc-ab-ac}{bc}+\frac{c^2+a^2+2ac-bc-ab}{ac}+\frac{a^2+b^2+2ab-bc-ab}{ab}$

$=\frac{a(b^2+c^2+2bc-ab-ac)+b(c^2+a^2+2ac-bc-ab)+c(a^2+b^2+2ab-bc-ac)}{abc}$

$=\frac{ab^2+ac^2+2abc-a^2b-a^2c+bc^2+a^2b+2abc-b^2c-ab^2+a^2c+b^2c+2abc-bc^2-ac^2}{abc}$

$=\frac{6abc}{abc}$                ∵ (All others will cut each other)

$=6$

Final answer:

$\frac{b+c(b+c-a)}{bc}+\frac{c+a(c+a-b)}{ca}+\frac{a+b(a+b-c)}{ab}=6$