If 1a-1a-b-1c-1c-b-0 and a-c-b- 0- then a- b- c satisfy-

The correct option is C b=2aca+c Given, 1a+1a b+1c+1c b=0 c b+aa(c b)+c+a bc(a b)=0 ac bc= ca+bc 2ac=bc+ac b=2aca+c ...

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Multiplying Terms with Same Base

If 1a+1a-b+...

Question

If $\frac{1}{a} + \frac{1}{a - b} + \frac{1}{c} + \frac{1}{c - b} = 0$ and $a + c - b \neq 0$ , then a, b, c satisfy.

2 b = a + c

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b^{2} = a c

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b = \frac{2 a c}{a + c}

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b^{2} = a^{2} + c^{2}

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Solution

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If a+b+c=0 then $(\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b})$ =?
(a) 1 (b) 0 (c) -1 (d) 3

a + b + c = 0

\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}

a, b, c

a + b - c = 1

a^{2} + b^{2} - c^{2} = - 1

a^{2} + b^{2} + c^{2}

a, b, c

x^{3} + q x + r = 0

(1) (b - c)^{2} + (c - a)^{2} + (a - b)^{2}

(2) (b + c)^{- 1} + (c + a)^{- 1} + (a + b)^{- 1}

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