Question

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Answer 1

Given :-

If a, b, c are positive real numbers such that

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

To find :-

• Value of $\sf \dfrac{(a+b)(b+c)(c+a)}{abc}$

Solution :-

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

Solve this step by step

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

$\implies\sf b(a+b-c)=c(a-b+c)$

$\implies\sf ba+b^2-cb=ca-bc+c^2$

$\implies\sf ba+b^2-cb+bc= ca+c^2$

$\implies\sf ba+b^2= ca+c^2$

$\implies\sf b^2-c^2=ca-ba$

$\implies\sf b^2-c^2= -a(-c+b)$

• Apply identity
• a² - b²=(a + b)(a - b)

$\\\implies\sf (b+a)(b-c)= -a(-c+b)$

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

$\therefore\sf{\underline{\boxed{\sf a=-(b+a)}}}$

Similarly

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

$\implies\sf c(-a+b-c)=a(a-b+c)$

$\implies\sf -ca+cb+c^2=a^2+ab-ac$

$\implies\sf -ca+ac+cb+c^2=a^2+ab$

$\implies\sf cb+c^2= a^2+ab$

$\implies\sf c^2-a^2=ab-cb$

$\implies\sf c^2-a^2= -b(-a+c)$

• Apply identity
• a² - b²=(a + b)(a - b)

$\\\implies\sf (c+a)(c-a)= -b(-a+c)$

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

$\therefore\sf{\underline{\boxed{\sf{b=-(c+a)}}}}$

Now

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

$\implies\sf a(a-b+c)=b(-a+b+c)$

$\implies\sf a^2-ab+ac=-ba+b^2+bc$

$\implies\sf a^2-ab+ba+ac=b^2+bc$

$\implies\sf a^2+ac= b^2+bc$

$\implies\sf a^2-b^2=bc-ac$

$\implies\sf a^2-b^2= -c(-b+c)$

• Apply identity
• a² - b²=(a + b)(a - b)

$\\\implies\sf (a+b)(a-b)=-c(-b+c)$

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

$\therefore\sf{\underline{\boxed{\sf{c=-(a+b)}}}}$

The value of

$\\\implies\sf \dfrac{(a+b)(b+c)(c+a)}{abc}$

Putting the value of a, b & c

$\\\implies\sf \dfrac{(a+b)(b+c)(c+a)}{[-(b+c)][-(c+a)][-(b+a)]}$

Cancel (a + b), (b + c) & (c + a)

$\\\implies\sf -1$

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Answer 2

Answer:

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Step-by-step explanation:

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