If a b-b c-c a-0- then the value of 1 - a2-b c-1 - b2-c a-1 - c2-a b wil be-A- 1B - a - b - cC-abcD-0

Given ab+bc+ca=0 and asked to find value 1/(a2 bc ) + 1/(c2 ab) + 1/(a2 bc) Put ac = ab + bc ; ab = ac+ bc and bc =ab + ac you get : =(1/(a2+ab+ac))+(1/(c2+ ...

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Byju's Answer

Standard VIII

Mathematics

Finding Square Root through Repeated Subtraction

If a b+b c+c ...

Question

If $a b + b c + c a = 0$ , then the value of $\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - c a} + \frac{1}{c^{2} - a b}$ will be:

A. $- 1$
B. $a + b + c$
C. $a b c$
D. $0$

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Solution

Correct option is D. $0$
Given: $a b + b c + c a = 0$
Therefore,
$\Rightarrow a b = - (b c + c a)$ ...(i)
$\Rightarrow b c = - (a b + c a)$ ...(ii) and
$\Rightarrow c a = - (a b + b c)$ ...9(ii)
Now,
From (i),(ii) and (iii)
$\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - c a} + \frac{1}{c^{2} - a b} = \frac{1}{a^{2} + (a b + c a)} + \frac{1}{b^{2} + (a b + b c)} + \frac{1}{c^{2} + (b c + c a)}$
$= \frac{1}{a (a + b + c)} + \frac{1}{b (a + b + c)} + \frac{1}{c (a + b + c)}$
$= \frac{b c + a c + a b}{a b c (a + b + c)}$
$= \frac{0}{a b c (a + b + c)}$
$= 0$
Therefore, $\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - c a} + \frac{1}{c^{2} - a b} = 0$

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If ab+bc+ca=0, then the value of 1a2−bc+1b2−ca+1c2−ab will be: A.−1 B.a+b+c C.abc D.0

If $a b + b c + c a = 0$ , then the value of $\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - c a} + \frac{1}{c^{2} - a b}$ will be:

A. $- 1$
B. $a + b + c$
C. $a b c$
D. $0$