If ab-bc-ca-0- then the value of 1a2-bc-1b2-ca-1c2-ab is -

The correct option is D 0 Put bc=ab+ca, ca=ab+bc, #160; ab=bc+ca , We get 1a(a+b+c)+1b(a+b+c)+1c(a+b+c) 1a+b+c(ab+bc+caabc) =0 ...

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Sum of n Terms

If ab+bc+ca...

Question

If $a b + b c + c a = 0$ , then the value of $\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - c a} + \frac{1}{c^{2} - a b}$ is ____.

- 1

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a + b + c

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a b c

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0

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Solution

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Similar questions

a + b + c = 0

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

(a b c \neq 0)

a b + b c + c a = 0

\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - c a} + \frac{1}{c^{2} - a b} = 0

a + b + c = 0

- (\frac{a^{2} + b^{2} + c^{2}}{b c + c a + a b})

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

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